0 of 81 Questions completed
Questions:
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading…
You must sign in or sign up to start the quiz.
You must first complete the following:
0 of 81 Questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 point(s), (0)
Earned Point(s): 0 of 0, (0)
0 Essay(s) Pending (Possible Point(s): 0)
Taking the Bohr radius as \(a_0=53 \mathrm{pm}\), the radius of \(\mathrm{Li}^{++}\)ion in its ground state, on the basis of Bohr’s model, will be about
(c) The atomic number \((Z)\) of lithium is 3.
As \(r_n=a_0 \frac{n^2}{Z}\),
Therefore, the radius of \(\mathrm{Li}^{++}\)ion in its ground state, on the basis of Bohr’s model, will be about \(\frac{1}{3}\) times to that of Bohr radius. Therefore, the radius of lithium ion is near \(r=\frac{53}{3} \approx 18 \mathrm{pm}\).
The binding energy of a H -atom, considering an electron moving around a fixed nuclei (proton), is \(B=-\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2}\). ( \(m=\) electron mass).
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving arround it. By similar arguments, the binding energy would be
\(
B=-\frac{M e^4}{8 n^2 \varepsilon_0^2 h^2}(M=\text { proton mass })
\)
This last expression is not correct because
(c) In a hydrogen atom, electrons revolving around a fixed proton nucleus have some centripetal acceleration. Therefore its frame of reference is non-inertial. If the frame of reference, where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference.
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because
(a) The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons because when we derive the formula for radius/energy levels etc, we make the assumption that centripetal force is provided only by electrostatic force of attraction by the nucleus. Hence, this will only work for single electron atoms. In multi-electron atoms, there will also be repulsion due to other electrons. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons.
For the ground state, the electron in the H -atom has an angular momentum \(=\hbar\), according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) Key concept: Bohr found that the magnitude of the electron’s angular momentum is quantized i.e. \(L=m v_n r_n=n\left(\frac{h}{2 \pi}\right)\) where \(n=1,2,3, \ldots\). each value of \(n\) corresponds to a permitted value of the orbit radius.
\(r_n=\) Radius of \(n^{\text {th }}\) orbit,\(v_n=\) corresponding speed
Bohr’s model gives only the magnitude of angular momentum. The angular momentum is a vector quantity. Hence we cannot express angular momentum completely by Bohr model. Hence it does not give correct values of angular momentum of revolving electron.
\(\mathrm{O}_2\) molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) Key concept: Forces that keep the nucleons bound in the nucleus are called nuclear forces.
Nuclear forces are short range forces. These do not exist at large distances greater than 10~15 m.
Nuclear forces are the strongest forces in nature.
These are attractive force and causes stability of the nucleus.
These forces are charge independent.
Nuclear forces arc non-central force.
The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. In \(\mathrm{O}_2\) molecule which consists of two oxygen atoms molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged and act inside the nucleus only.
Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(a) Let two H atoms initially at in the ground state. Now two atoms collide inelastically. The total energy associated with the two H atoms
\(
=2 \times(13.6 \mathrm{eV})=27.2 \mathrm{eV}
\)
The maximum amount by which their combined kinetic energy is reduced when any one of them goes into first excited state \((n=2)\) after the inclastic collision.
The total energy associated with the two H-atoms after the collision
\(
=\left(\frac{13.6}{2^2}\right)+(13.6)=17.0 \mathrm{eV}
\)
Hence, maximum loss of their combined kinetic energy
\(
=27.2-17.0=10.2 \mathrm{eV}
\)
A set of atoms in an excited state decays.
(a) When atoms in an excited state decay, they generally transition to a lower energy state, often emitting energy in the form of photons, and this process is known as de-excitation.
An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,
(a, c) In a hydrogen atom, electron revolves around a fixed proton nucleus in circular path. This can be explained by Bohr model. But in case of ionised H -molecule which consists of two protons in nucleus and where protons are separated by a small distance of the order of angstrom, cannot be explained by Bohr model. Hence in this case the ground state the electron would not move in circular orbits, the electrons orbit would go around the protons.
Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,
(a, b) A moving electron and proton cannot combine to produce a H -atom because of energy conservation and without simultaneously releasing energy in the form of radiation.
The Bohr model for the spectra of a H-atom
(a, b) The model of the atom proposed by Neil Bohr was applicable for the hydrogen atom and some other light atoms containing only one electron around the nucleus which is stationary having a positive charge, Ze (called a hydrogen-like atom, \(\text { e.g.: } \left.\mathrm{H}, \mathrm{He}^{+}, \mathrm{Li}^{+2}, \mathrm{Na}^{+1} \mathrm{etc}\right)\). But one of the limitations of the model is that it is not applicable to hydrogen in its molecular form and also in case of He atom.
The Balmer series for the H -atom can be observed
(b, d) The Balmer series for the H -atom can be observed if we measure the frequencies of light emitted due to transitions between excited states and the first excited state and as a sequences of frequencies with the higher frequencies getting closely packed.
Let \(E_n=\frac{-1}{8 \varepsilon_0^2} \frac{m e^4}{n^2 h^2}\) be the energy of the \(n^{\text {th }}\) level of H -atom. If all the H -atoms are in the ground state and radiation of frequency \(\left(E_2-E_1\right) / h\) falls on it,
(b, d) Let \(\mathrm{E}_2[latex] and [latex]\mathrm{E}_1\) be the energy corresponding to \({n}=2\) and \({n}=1\) respectively. If radiation of energy \(\Delta E=\left(E_2-E_1\right)=hf\) incident on a sample where all the H -atoms are in the ground state, according to Bohr model some of the atoms will move to the first excited state. As this energy is not sufficient for transition from \(n=1\) to \(n=3\), hence no atoms will make a transition to the \(n=3\) state.
The simple Bohr model is not applicable to \(\mathrm{He}^4\) atom because
(c, d) Key concept: Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge \(Ze\) (called hydrogen like atom). It is valid only for one electron atoms, e.g. : \(\mathrm{H}, \mathrm{He}^{+}, \mathrm{Li}^{+2}, \mathrm{Na}^{+1}\) etc.
The Bohr model is not applicable to \(\)\mathrm{He}^4\(\), as it has one more electron and electrons are not subjected to central forces.
Calculate the energy of a \(\mathrm{He}^{+}\)ion in its first excited state.
(b) The energy is \(E_n=\frac{-R h c Z^2}{n^2}=-\frac{(13.6 \mathrm{eV}) Z^2}{n^2}\)
For a \(\mathrm{He}^{+}\)ion, \(Z=2\) and for the first excited state, \(n\) \(=2\) so that the energy of \(\mathrm{He}^{+}\)ion in the first excited state is -13.6 eV.
The excitation energy of a hydrogen-like ion in its first excited state is 40.8 eV. Find the energy needed to remove the electron from the ion.
(a) The excitation energy in the first excited state is
\(
\begin{aligned}
E & =R h c Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\
& =(13.6 \mathrm{eV}) \times Z^2 \times \frac{3}{4}
\end{aligned}
\)
Equating this to 40.8 eV , we get \(Z=2\). So, the ion in question is \(\mathrm{He}^{+}\). The energy of the ion in the ground state is
\(
\begin{aligned}
E & =-\frac{R h c Z^2}{1^2}=-4 \times(13.6 \mathrm{eV}) \\
& =-54.4 \mathrm{eV}
\end{aligned}
\)
Thus 54.4 eV is required to remove the electron from the ion.
Find the radius of \(\mathrm{Li}^{++}\)ions in its ground state assuming Bohr’s model to be valid.
(c) For hydrogen-like ions, the radius of the \(n\)th orbit is
\(
a_n=\frac{n^2 a_0}{Z}
\)
For \(\mathrm{Li}^{++}, Z=3\) and in ground state \(n=1\). The radius is
\(
a_1=\frac{53 \mathrm{pm}}{3} \approx 18 \mathrm{pm} \text { First Bohr radius } a_0=53 \mathrm{pm}
\)
A particular hydrogen-like ion emits radiation of frequency \(2.467 \times 10^{15} \mathrm{~Hz}\) when it makes transition from \(n=2\) to \(n=1\). What will be the frequency of the radiation emitted in a transition from \(n=3\) to \(n=1\)?
(b) The frequency of radiation emitted is given by
\(
\nu=\frac{c}{\lambda}=K\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
Thus, \(2.467 \times 10^{15} \mathrm{~Hz}=K\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)
\(
K=\frac{4}{3} \times 2.467 \times 10^{15} \mathrm{~Hz}
\)
The frequency of the radiation emitted in the transition \(n=3\) to \(n=1\) is
\(
\begin{aligned}
\nu^{\prime} & =K\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \\
& =\frac{8}{9} K=\frac{8}{9} \times \frac{4}{3} \times 2.467 \times 10^{15} \mathrm{~Hz} \\
& =2.92 \times 10^{15} \mathrm{~Hz}
\end{aligned}
\)
Calculate the two highest wavelengths of the radiation emitted when hydrogen atoms make transitions from higher states to \(n=2\) states.
(a) The highest wavelength corresponds to the lowest energy of transition. This will be the case for the transition \(n=3\) to \(n=2\). The second highest wavelength corresponds to the transition \(n=4\) to \(n=2\).
The energy of the state \(n\) is \(E_n=\frac{E_1}{n^2}\).
Thus,
\(
\begin{aligned}
& E_2=-\frac{13.6 \mathrm{eV}}{4}=-3.4 \mathrm{eV} \\
& E_3=-\frac{13.6 \mathrm{eV}}{9}=-1.5 \mathrm{eV}
\end{aligned}
\)
and \(E_4=-\frac{13.6 \mathrm{eV}}{16}=-0.85 \mathrm{eV}\)
The highest wavelength is \(\lambda_1=\frac{h c}{\Delta E}\)
\(
=\frac{1242 \mathrm{eVnm}}{(3 \cdot 4 \mathrm{eV}-1 \cdot 5 \mathrm{eV})}=654 \mathrm{~nm}
\)
The second highest wavelength is
\(
\lambda_2=\frac{1242 \mathrm{eVnm}}{(3 \cdot 4 \mathrm{eV}-0 \cdot 85 \mathrm{eV})}=487 \mathrm{~nm}
\)
What is the wavelength of the radiation emitted when the electron in a hydrogen atom jumps from \(n=\infty\) to \(n=2\)?
(d) The energy of \(n=2\) state is
\(
E_2=\frac{-13.6 \mathrm{eV}}{4}=-3.4 \mathrm{eV}
\)
The energy of \(n=\infty\) state is zero.
\(
\begin{aligned}
&\text { The wavelength emitted in the given transition is }\\
&\begin{aligned}
\lambda & =\frac{h c}{\Delta E} \\
& =\frac{1242 \mathrm{eV} \mathrm{~nm}}{3 \cdot 4 \mathrm{eV}}=365 \mathrm{~nm}
\end{aligned}
\end{aligned}
\)
How many different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number \(n\)?
(c) From the \(n\)th state, the atom may go to \((n-1)\) th state, \(\ldots, 2\) nd state or 1st state. So there are \((n-1)\) possible transitions starting from the \(n\)th state.
The atoms reaching \((n-1)\) th state may make \((n-2)\) different transitions. Similarly for other lower states. The total number of possible transitions is
\(
\begin{aligned}
& (n-1)+(n-2)+(n-3)+\ldots 2+1 \\
& =\frac{n(n-1)}{2}
\end{aligned}
\)
Monochromatic radiation of wavelength \(\lambda\) is incident on a hydrogen sample in the ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. Find the value of \(\lambda\).
(a) As the hydrogen atoms emit radiation of six different wavelengths, some of them must have been excited to \(n=4\). The energy in \(n=4\) state is
\(
E_4=\frac{E_1}{4^2}=-\frac{13.6 \mathrm{eV}}{16}=-0.85 \mathrm{eV} .
\)
The energy needed to take a hydrogen atom from its ground state to \(n=4\) is
\(
13.6 \mathrm{eV}-0.85 \mathrm{eV}=12.75 \mathrm{eV}
\)
The photons of the incident radiation should have 12.75 eV of energy. So
\(
\begin{aligned}
\frac{h c}{\lambda} & =12.75 \mathrm{eV} \\
\lambda & =\frac{h c}{12.75 \mathrm{eV}} \\
& =\frac{1242 \mathrm{eV} \mathrm{~nm}}{12.75 \mathrm{eV}}=97.5 \mathrm{~nm} .
\end{aligned}
\)
The energy needed to detach the electron of a hydrogen-like ion in ground state is 4 rydberg. (i) What is the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state ? (ii) What is the radius of the first orbit for this atom?
(b) (i) In energy units, 1 rydberg \(=13.6 \mathrm{eV}\). The energy needed to detach the electron is \(4 \times 13.6 \mathrm{eV}\). The energy in the ground state is, therefore, \(E_1=-4 \times 13.6 \mathrm{eV}\). The energy of the first excited state \((n=2)\) is \(E_2=\frac{E_1}{4}=-13.6 \mathrm{eV}\). The energy difference is \(E_2-E_1=3 \times 13.6 \mathrm{eV}=40.8 \mathrm{eV}\). The wavelength of the radiation emitted is
\(
\begin{aligned}
\lambda & =\frac{h c}{\Delta E} \\
& =\frac{1242 \mathrm{eV} \mathrm{~nm}}{40 \cdot 8 \mathrm{eV}}=30 \cdot 4 \mathrm{~nm}
\end{aligned}
\)
(ii) The energy of a hydrogen-like ion in ground state is \(E=Z^2 E_0\) where \(Z=\) atomic number and \(E_0=-13.6 \mathrm{eV}\). Thus, \(Z=2\). The radius of the first orbit is \(\frac{a_0}{Z}\) where \(a_0=53 \mathrm{pm}\). Thus,
\(
r=\frac{53 \mathrm{pm}}{2}=26 \cdot 5 \mathrm{pm}
\)
A hydrogen sample is prepared in a particular excited state A. Photons of energy 2.55 eV get absorbed into the sample to take some of the electrons to a further excited state \(B\). Find the quantum numbers of the states \(A\) and \(B\).
(d) The allowed energies of hydrogen atoms are
\(
\begin{aligned}
& E_1=-13.6 \mathrm{eV} \\
& E_2=-3.4 \mathrm{eV} \\
& E_3=-1.5 \mathrm{eV} \\
& E_4=-0.85 \mathrm{eV} \\
& E_5=-0.54 \mathrm{eV}
\end{aligned}
\)
We see that a difference of 2.55 eV can only be absorbed in transition \(n=2\) to \(n=4\). So the state \(A\) has quantum number 2 and the state \(B\) has quantum number 4.
(i) Find the maximum wavelength \(\lambda_0\) of light which can ionize a hydrogen atom in its ground state. (ii) Light of wavelength \(\lambda_0\) is incident on a hydrogen atom which is in its first excited state. Find the kinetic energy of the electron coming out.
(a) (i) To ionize a hydrogen atom in ground state, a minimum of 13.6 eV energy should be given to it. A photon of light should have this much of energy in order to ionize a hydrogen atom. Thus,
\(
\frac{h c}{\lambda_0}=13.6 \mathrm{eV}
\)
\(
\lambda_0=\frac{1242 \mathrm{eV} \mathrm{~nm}}{13 \cdot 6 \mathrm{eV}}=91 \cdot 3 \mathrm{~nm} .
\)
(ii) The energy of the hydrogen atom in its first excited state is \(-\frac{13.6 \mathrm{eV}}{4}=-3.4 \mathrm{eV}\). Thus, 3.4 eV of energy is needed to take the electron out of the atom. The energy of a photon of the light of wavelength \(\lambda_0\) is 13.6 eV. Thus, the electron coming out will have a kinetic energy \(13.6 \mathrm{eV}-3.4 \mathrm{eV}=10.2 \mathrm{eV}\).
A lithium atom has three electrons. Assume the following simple picture of the atom. Two electrons move close to the nucleus making up a spherical cloud around it and the third moves outside this cloud in a circular orbit. Bohr’s model can be used for the motion of this third electron but \(n=1\) states are not available to it. Calculate the ionization energy of lithium in ground state using the above picture.
(a) In this picture, the third electron moves in the field of a total charge \(+3 e-2 e=+e\). Thus, the energies are the same as that of hydrogen atoms. The lowest energy is
\(
E_2=\frac{E_1}{4}=\frac{-13.6 \mathrm{eV}}{4}=-3.4 \mathrm{eV}
\)
Thus, the ionization energy of the atom in this picture is 3.4 eV.
Light corresponding to the transition \(n=4\) to \(n=2\) in hydrogen atoms falls on cesium metal (work function \(=1.9 \mathrm{eV}\) ). Find the maximum kinetic energy of the photoelectrons emitted.
(c) The energy of the photons emitted in transition \(n=4\) to \(n=2\) is
\(
h \nu=13.6 \mathrm{eV}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=2.55 \mathrm{eV}
\)
The maximum kinetic energy of the photoelectrons is \(=2.55 \mathrm{eV}-1.9 \mathrm{eV}=0.65 \mathrm{eV}\).
The minimum orbital angular momentum of the electron in a hydrogen atom is
(c) According to Bohr’s atomic theory, the orbital angular momentum \(L\) of an electron is an integral multiplt of \(h / 2 \pi\).
\(
\therefore L=\frac{n h}{2 \pi}
\)
Here,
\(n=\) Principal quantum number
The minimum value of \(n=\) is 1.
Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by
\(
L=\frac{h}{2 \pi}
\)
Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV and 1.9 eV. These photons must come from
(d) The energies of the photons emitted can be expressed as follows:
\(
\begin{aligned}
& 13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \\
& 13.6\left(\frac{1}{1^2}-\frac{1}{3^2}\right) \mathrm{eV}=12.1 \mathrm{eV} \\
& 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \mathrm{eV}=1.9 \mathrm{eV}
\end{aligned}
\)
The following table gives the transition corresponding to the energy of the photon:
\(
\begin{array}{|l|l|}
\hline \text { Energy of photon } & \text { Transition } \\
\hline 12.1 \mathrm{eV} & n=3 \text { to } n=1 \\
\hline 10.2 \mathrm{eV} & n=2 \text { to } n=1 \\
\hline 1.9 \mathrm{eV} & n=3 \text { to } n=2 \\
\hline
\end{array}
\)
A hydrogen atom consists of only one electron. An electron can have transitions, like from \(n=3\) to \(n=2\) or from \(n=2\) to \(n=1\), at a time.
So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has \(n=3\) to \(n=2\) and then \(n=2\) to \(n=1\) electronic transition and the other has \(n=3\) to \(n=1\) electronic transition).
Suppose, the electron in a hydrogen atom makes transition from \(n=3\) to \(n=2\) in \(10^{-8} \mathrm{~s}\). The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is
(c) The angular momentum of the electron for the \(n\)th state is given by
\(
L_n=\frac{n h}{2 \pi}
\)
Angular momentum of the electron for \(n=3\),
\(
\mathrm{n}=3, L_i=\frac{3 h}{2 \pi}
\)
Angular momentum of the electron for \(n=2, L_f=\frac{2 h}{2 \pi}\)
The torque is the time rate of change of the angular momentum.
\(
\begin{aligned}
& \text { Torque } \tau=\frac{L_f-L_i}{t} \\
& =\frac{(2 h / 2 \pi)-(3 h / 2 \pi)}{10^{-8}} \\
& =-\frac{h / 2 \pi}{\left(10^{-8}\right)} \\
& =\frac{-10^{-34}}{10^{-8}} \cdots \cdots \cdots \cdots \cdot\left[\therefore \frac{h}{2 \pi} \approx 10^{-34} \mathrm{~J}-\mathrm{s}\right] \\
& =-10^{-42} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)
The magnitude of the torque is \(10^{-42} N m\)
In which of the following transitions will the wavelength be minimum?
(d) For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by
\(
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n^1}-\frac{1}{n^2}\right)
\)
Here, \(R\) is the Rydberg constant.
For the transition from \(n=5\) to \(n=4\), the wavelength is given by
\(
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left(\frac{1}{4^2}-\frac{1}{5^2}\right) \\
& \lambda=\frac{400}{9 R Z^2}
\end{aligned}
\)
For the transition from \(n=4\) to \(n=3\), the wavelength is given by
\(
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\
& \lambda=\frac{144}{7 R Z^2}
\end{aligned}
\)
For the transition from \(n=3\) to \(n=2\), the wavelength is given by
\(
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \lambda=\frac{36}{5 R Z^2}
\end{aligned}
\)
For the transition from \(n=2\) to \(n=1\), the wavelength is given by
\(
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \lambda=\frac{2}{R Z^2}
\end{aligned}
\)
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from \(n=2\) to \(n=1\) will be minimum.
In which of the following systems will the radius of the first orbit ( \(n=1\) ) be minimum?
(d) For a hydrogen-like ion with \(Z\) protons in the nucleus, the radius of the \(n\)th state is given by
\(
r_n=\frac{n^2 a_0}{Z}
\)
Here , \(a_0=0.53 \mathrm{pm}\)
For lithium,
\(
Z=3
\)
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.
Explanation:
According to Bohr’s model, the radius of an orbit is inversely proportional to the atomic number \((Z)\) of the atom.
Hydrogen: \(\mathrm{Z}=1\)
Deuterium: \(Z=1\) (Deuterium is an isotope of hydrogen)
Singly ionized helium: \(Z=2\)
Doubly ionized lithium: \(Z=3\)
Since a higher atomic number means a smaller radius, the doubly ionized lithium will have the smallest radius for the first orbit.
In which of the following systems will the wavelength corresponding to \(n=2\) to \(n=1\) be minimum?
(d) The wavelength corresponding the transition from \(n_2\) to \(n_1\) is given by
\(
\left.\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\right]
\)
Here,
\(R=\) Rydberg constant
\(Z=\) Atomic number of the ion
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to \(n=2\) to \(n=1\) will be minimum in doubly ionized lithium ion because for lithium, \(Z=3\).
Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number \(n\)?
(c) The speed \((v)\) of electron can be expressed as
\(
v=\frac{Z e^2}{2 \in_0 h n} \dots(1)
\)
Here,
\(Z=\) Number of protons in the nucleus
\(e=\) Magnitude of charge on electron charge
\(n=\) Principal quantum number
\(h=\) Planck’s constant
It can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number ( \(n\) ).
Therefore, the graph between them must be a rectangular hyperbola.
As one considers orbits with higher values of \(n\) in a hydrogen atom, the electric potential energy of the atom
(b) In a hydrogen atom, the total energy \((E)\) is related to the potential energy by the equation:
\(
P E=2 E
\)
The total energy for a hydrogen atom can be expressed as:
\(
E=-\frac{13.6 Z^2}{n^2} \mathrm{eV}
\)
For hydrogen, \(Z=1\), so:
\(
E=-\frac{13.6}{n^2} \mathrm{eV}
\)
Calculating Potential Energy:
Substituting the expression for total energy into the potential energy equation:
\(
P E=2 E=2\left(-\frac{13.6}{n^2}\right)=-\frac{27.2}{n^2} \mathrm{eV}
\)
Analyzing the Effect of Increasing \(\boldsymbol{n}\) :
As \(n\) increases, the term \(n^2\) in the denominator increases, which means that the absolute value of the potential energy (which is negative) becomes smaller.
Thus, the potential energy becomes less negative:
\(P E\) increases (less negative) as \(n\) increases.
The energy of an atom (or ion) in its ground state is -54.4 eV. It may be
(c) The total energy of a hydrogen-like ion, having \(Z\) protons in its nucleus, is given by
\(
E=-\frac{13.6 Z^2}{n^2} e V
\)
Here, \(n=\) Principal quantum number
For ground state,
\(
n=1
\)
\(\therefore\) Total energy, \(E=-13.6 Z^2 \mathrm{eV}\)
For hydrogen,
\(Z=1\)
\(\therefore\) Total energy, \(E=-13.6 \mathrm{eV}\)
For deuterium,
\(
Z=1
\)
\(\therefore\) Total energy, \(E=-13.6 \mathrm{eV}\)
For \(\mathrm{He}^{+}\),
\(
Z=2
\)
\(\therefore\) Total energy, \(E=-13.6 \times 2^2=-54.4 \mathrm{eV}\)
For \(\mathrm{Li}^{++}\),
\(
Z=3
\)
\(\therefore\) Total energy, \(E=-13.6 \times 3^2=-122.4 \mathrm{eV}\)
Hence, the ion having an energy of -54.4 eV in its ground state may be \(\mathrm{He}^{+}\).
The radius of the shortest orbit in a one-electron system is 18 pm. It may be
(d) The radius of the \(n[latex]th orbit in one electron system is given by
[latex]
r_n=\frac{n^2 a_0}{Z}
\)
Here, \(a_0=53 \mathrm{pm}\)
For the shortest orbit,
\(
n=1
\)
For hydrogen,
\(
Z=1
\)
\(\therefore\) Radius of the first state of hydrogen atom \(=53 \mathrm{pm}\)
For \(\mathrm{He}^{+}\),
\(Z=2\)
\(\therefore\) Radius of \(\mathrm{He}^{+}\)atom \(=\frac{53}{2} \pm=26.5 \mathrm{pm}\)
For \(\mathrm{Li}^{++}\),
\(Z=3\)
\(\therefore\) Radius of \(\mathrm{Li}^{++}\)atom \(=\frac{53}{3} \mathrm{pm}=17.66 \mathrm{pm} \approx 18 \mathrm{pm}\)
The given one-electron system having radius of the shortest orbit to be 18 pm may be \(\mathrm{Li}^{++}\).
A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
(a) Let after absorption of energy, the hydrogen atom goes to the \(n\)th excited state.
Therefore, the energy absorbed can be written as
\(
\begin{aligned}
& 10.2=13.6 \times\left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\
& \Rightarrow \frac{10.2}{13.6}=1-\frac{1}{n^2} \\
& \Rightarrow=\frac{1}{n^2}=\frac{13.6-10.2}{13.6} \\
& \Rightarrow=\frac{1}{n^2}=\frac{3.4}{13.6} \\
& \Rightarrow n^2=4 \\
& \Rightarrow n=2
\end{aligned}
\)
The orbital angular momentum of the electron in the \(n\)th state is given by
\(
L_n=\frac{n h}{2 \pi}
\)
Change in the angular momentum,
\(
\begin{aligned}
& \Delta L=\frac{2 h}{2 \pi}-\frac{h}{2 \pi}=\frac{h}{2 \pi} \\
& \Delta L=1.05 \times 10^{-34} \mathrm{Js}
\end{aligned}
\)
Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?
(d) The parameter that is the same for all hydrogen-like atoms and ions in their ground states is Orbital angular momentum of the electron.
The orbital angular momentum of the electron is given by
\(
L=n \frac{h}{2 \pi}
\)
For the ground state, \(n=1\) :
\(
L=\frac{h}{2 \pi}
\)
This value is constant and does not depend on \(Z\). Therefore, the orbital angular momentum is the same for all hydrogen-like atoms in their ground state. The only parameter that is the same for all hydrogen-like atoms and ions in their ground state is the orbital angular momentum of the electron.
In a laser tube, all the photons
(d) All the photons emitted in the laser move with the speed equal to the speed of light \(\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)\).
Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.
In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory
(b) The primary reason for the difference in the observed hydrogen spectra is the temperature difference between a laboratory discharge tube and a star. In a laboratory setting, the hydrogen gas is at a much lower temperature compared to a star. This lower temperature means that only a small number of energy levels in the hydrogen atoms are excited, leading to a limited number of emission lines. In contrast, the high temperature of a star excites a much wider range of energy levels, resulting in a richer spectrum with many more lines, according to one explanation.
An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision
(a) In hydrogen atom \(E_2-E_1=10.2 \mathrm{eV}\)
Since, \(\quad 5 \mathrm{eV}<10.2 \mathrm{eV}\)
The incident electron has 5eV of kinetic energy, which is less than the 10.2eV required for excitation.
Elastic Collision: Since the electron doesn’t have enough energy to excite the atom, it can only bounce off elastically. This means it retains most of its kinetic energy.
Which of the following products in a hydrogen atom are independent of the principal quantum number \(n\)? The symbols have their usual meanings.
(a, b) Relations for energy, radius of the orbit and its velocity are given by
\(
\begin{aligned}
& E=-\frac{m Z^2 e^4}{8 \in_0^2 h^2 n^2} \\
& r=\frac{\in_0 h^2 n^2}{\pi m Z e^2} \\
& v=\frac{Z e^2}{2 \in_0 h n}
\end{aligned}
\)
Where
\(Z\) : the atomic number of hydrogen like atom
\(e\) : electric charge
\(h\) : plank constant
\(m\) : mass of electron
\(n\) : principal quantam number of the electron
\(\in_0\) : permittivity of vacuum
From these relations, we can see that the products independent of \(n\) are \(v n, \mathrm{E} r\).
Let \(A_n\) be the area enclosed by the \(n\)th orbit in a hydrogen atom. The graph of \(\ln \left(A_n / A_1\right)\) against \(\ln (n)\)
(a, b) The radius of the \(n\)th orbit of a hydrogen atom is given by
\(
r_n=n^2 a_0
\)
Area of the nth orbit is given by
\(
\begin{aligned}
& A_n=\pi r_n^2=\pi n^4 a_0^2 \\
& A_1=\pi a_0^2 \\
& \Rightarrow \operatorname{In}\left(\frac{A n}{A_1}\right)=\ln \left(\frac{\pi n^4 a_0^2}{\pi a_0^2}\right) \\
& \operatorname{In}\left(\frac{A n}{A_1}\right)=4 \ln n \ldots \ldots \ldots(1)
\end{aligned}
\)
From the above expression, the graph of \(\ln \left(\mathrm{A}_n / \mathrm{A}_1\right)\) against \(\ln (n)\) will be a straight line passing through the origin and having slope 4.
Ionization energy of a hydrogen-like ion \(A\) is greater than that of another hydrogen-like ion \(B\). Let \(r, u, E\) and \(L\) represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state
(b) The ionisation energy of a hydrogen like ion of atomic number \(Z\) is given by
\(
V=(13.6 \mathrm{eV}) \times Z^2
\)
Thus, the atomic number of ion A is greater than that of \(\mathrm{B}\left(Z_{\mathrm{A}}>Z_{\mathrm{B}}\right)\).
The radius of the orbit is inversely proportional to the atomic number of the ion.
\(
\therefore r_{\mathrm{A}}>r_{\mathrm{B}}
\)
Thus, (a) is incorrect.
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of A will be more than that in B.
Thus, \(u_{\mathrm{A}}>u_{\mathrm{B}}\) is correct.
The total energy of the atom is given by
\(
E=-\frac{m Z^2 e^2}{8 \epsilon_0 h^2 n^2}
\)
As the energy is directly proportional to \(Z^2\), the energy of A will be less than that of B , i.e. \(E_{\mathrm{A}}<E_{\mathrm{B}}\).
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation \(L_{\mathrm{A}}>L_{\mathrm{B}}\) is invalid.
When a photon stimulates the emission of another photon, the two photons have
(a, b, c, d) When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent.
When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic and unidirectional light is called lasing action.
Using known values for hydrogen atom, calculate speed of electron in fourth orbit of \(\mathrm{He}^{+}\).
(b) As, for \(\mathrm{He}^{+}(Z=2)\),
Also, we know that, \(v_n=\frac{Z}{n} v_1\)
Substituting \(n=4, Z=2\) and \(v_1=2.2 \times 10^6 \mathrm{~ms}^{-1}\), we get
\(
v_4=\left(\frac{2}{4}\right)\left(2.2 \times 10^6\right)=1.1 \times 10^6 \mathrm{~ms}^{-1}
\)
\(
\begin{aligned}
&\text { Velocity of electron in } n \text {th orbit is, }\\
&v_n=2.2 \times 10^6 \frac{Z}{n} \mathrm{~ms}^{-1}=\frac{Z}{n} v_1
\end{aligned}
\)
Find the ratio of product of velocity and time period of electron orbiting in 2nd and 3rd stable orbits for an hydrogen like atom.
(c) We know that, time period \(T_n \propto n^3\) and velocity \(v_n \propto 1 / n\).
\(
\begin{aligned}
& \Rightarrow \text { Time period }\left(T_n\right) \times \text { velocity }\left(v_n\right) \propto n^2 \\
& \therefore \quad \frac{T_2 v_2}{T_3 v_3}=\frac{2^2}{3^2}=\frac{4}{9}
\end{aligned}
\)
In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius \(5 \times 10^{-11} \mathrm{~m}\) at a speed \(2.2 \times 10^6 \mathrm{~ms}^{-1}\). Calculate the current associated with electron motion?
(d) Frequency of revolution of electron can be written as,
\(
\nu=\frac{v}{2 \pi r}=\frac{2.2 \times 10^6}{2 \pi\left(5 \times 10^{-11}\right)}=7 \times 10^{15} \mathrm{~Hz}
\)
\(\therefore\) Current associated, \(I=q / t=q \nu\)
\(
\left(\because \frac{1}{t}=\nu\right)
\)
\(
=\left(1.6 \times 10^{-19}\right)\left(7 \times 10^{15}\right)=11.2 \times 10^{-4} \mathrm{~A}=1.12 \mathrm{~mA}
\)
The radius of an atom is of the order of
(d) \(\text { The radius of an atom is typically described as being on the order of } 10^{-10} \mathrm{~m} \)
An alpha nucleus of energy \(\frac{1}{2} m v^2\) bombards a heavy nuclear target of charge \(Z e\). Then, the distance of closest approach for the alpha nucleus will be proportional to
\(
\begin{aligned}
&\text { (b) For the distance of closest approach, we can write }\\
&\frac{1}{2} m v^2=\frac{k \times(Z e)(2 e)}{r_0} \Rightarrow r_0 \propto \frac{1}{m}
\end{aligned}
\)
The number of \(\alpha\)-particles scattered per unit area \(N(\theta)\) at scattering angle \(\theta\) varies inversely as
(b) The number of \(\alpha[latex]-particles scattered per unit area [latex]N(\theta)\) at a scattering angle \(\theta\) varies inversely as \(\sin ^4\left(\frac{\theta}{2}\right)\).
Explanation:
According to Rutherford’s scattering formula, the number of \(\alpha\)-particles scattered at an angle \(\theta\) is given by:
\(
N(\theta) \propto \frac{1}{\sin ^4\left(\frac{\theta}{2}\right)}
\)
This relationship indicates an inverse proportionality between the scattering intensity and the fourth power of the sine of half the scattering angle.
In Rutherford scattering experiment for scattering angle of \(360^{\circ}\), what will be the value of impact parameter?
\(
\text { (b) We know that, impact parameter, } b=K \cot \frac{\theta}{2}
\)
\(
\begin{array}{ll}
\text { Given, } & \theta=360^{\circ} \\
\therefore & b=K \times \cot 180^{\circ}=\infty
\end{array}
\)
Bohr’s atom model assumes
(d) Bohr’s Model: Niels Bohr by making \(\mathbf{4}\) postulates solved the puzzle of hydrogen spectra.
The mass of the nucleus is very large compared to that of the electrons and almost the entire mass of the atom is concentrated in the nucleus and hence assumed to be infinite.
The electrons revolve around the nucleus in circular orbits.
The mass of the electron remains constant.
The radius around the orbit has some special values of the radius. In these stationary orbits, they do not radiate energy as expected from Maxwell’s laws.
The energy of each stationary orbits are fixed, electrons can jump from a higher orbit to a lower orbit by emitting a photon of radiation. Where Higher energy orbit – Lower energy orbit = \((h c)/\) \(\lambda\)
An electron can also jump from lower to higher energy by absorbing energy.
In stationary orbits, the angular momentum \(\mathbf{L}\) of an electron is an integral multiple of ( \(\mathbf{h} / \mathbf{2} \boldsymbol{\pi}\) )
\(L=n \times(h / 2 \pi)\)
The angular momentum \((L)\) of an electron moving in a stable orbit around nucleus is
(c) For an electron to remain orbiting around the nucleus, the. angular momentum ( \(L\) ) should be an integral multiple of \(h / 2 \pi\).
ie, \(m v r=\frac{n h}{2 \pi}\) where \(n=\) principle quantum number of electron, and \(h=\) Planck’s constant
In Bohr’s model, the atomic radius of the first orbit is \(r_0\), then the radius of the third orbit is
\(
\begin{aligned}
& \text { (c) As, } r \propto n^2 \\
& \qquad \frac{r_3}{r_1}=\frac{3^2}{1^2}=9 \\
& \Rightarrow \quad r_3=9 r_1=9 r_0 \left(\because r_1=r_0\right)
\end{aligned}
\)
The ratio of the speed of the electrons in the ground state of hydrogen atom to the speed of light in vacuum is
(c) \(v_n=\frac{Z e^2}{2 h \varepsilon_0} \frac{1}{n}\)
The velocity of electron in the first orbit ( \(n=1\) ) of H -atom \((Z=1)\) is
\(
\begin{array}{ll}
& v_1=\frac{e^2}{2 h \varepsilon_0}=\frac{c}{137} \\
\therefore & \frac{v_1}{c}=\frac{1}{137}
\end{array}
\)
Speed \(\left(v_n\right)\) of an electron in \(n\)th orbit versus principal quantum number ( \(n\) ) graph is
(c) The speed of an electron in the nth orbit is given by the formula:
\(
V_n \propto \frac{Z}{n}
\)
where \(Z\) is the atomic number of the element. For a given atom, \(Z\) remains constant.
As \(n\) increases, \(V_n\) decreases.
Based on the graphical representation of the relationship \(V_n \propto \frac{1}{n}\), we can conclude that the correct option for the graph of \(V_n\) versus \(n\) is the one that shows a hyperbolic decline.
The orbital frequency of an electron in hydrogen like atom is proportional to
(b) Time period of revolution of electron in \(n\)th orbit,
\(
T_n=\frac{n h r_n}{k Z e^2} \left(\text { where, } k=\frac{1}{4 \pi \varepsilon_0}\right)
\)
Since, \(r_n \propto \frac{n^2}{Z}\)
\(
\therefore \quad T_n \propto n^3 \Rightarrow \frac{1}{\mathrm{v}_n} \propto n^3 \text { or } \mathrm{v}_n \propto n^{-3}
\)
The ground state energy of hydrogen atom is -13.6 eV. What is the kinetic and potential energies of the electron in this state?
(c) As, we know that, kinetic energy \(=-\) total energy
\(
=-(-13.6) \mathrm{eV}=13.6 \mathrm{eV}
\)
and potential energy \(=+2\) (total energy)
\(
=2(-13.6) \mathrm{eV}=-27.2 \mathrm{eV}
\)
The total energy of electron in the ground state of hydrogen atom is -13.6 eV. Find the kinetic energy of an electron in the first excited state.
(d) The energy of electron in hydrogen atom when it revolves in \(n\)th orbit,
\(
E_n=\frac{-13.6}{n^2} \mathrm{eV}
\)
In the ground state \((n=1)\),
\(
\therefore \quad E=\frac{-13.6}{1^2}=-13.6 \mathrm{eV}
\)
For first excited state \((n=2), E=\frac{-13.6}{2^2}=-3.4 \mathrm{eV}\)
So, kinetic energy of electron in the first excited state, i.e. for \((n=2)\), kinetic energy \(=-\) total energy
\(
K=-E=-(-3.4)=3.4 \mathrm{eV}
\)
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
(a) Here, difference of energy,
\(
\begin{aligned}
\Delta E=2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \mathrm{~J} & \\
& \left(\because 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)
\end{aligned}
\)
Using the relation, \(h \nu=\Delta E=E_2-E_1\)
\(
\begin{aligned}
\nu & =\frac{\Delta E}{h}=\frac{2.3 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}} \\
& =0.5559 \times 10^{15} \approx 5.6 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\)
Determine the wavelength of the radiation required to excite the electron in \(\mathrm{Li}^{++}\)from the first to the third Bohr orbit.
(d) Given, for \(\mathrm{Li}^{++}, Z=3\) and as the excitation is from first to third Bohr orbit, so \(n_1=1, n_2=3\)
Using the relation,
\(
\begin{aligned}
\frac{1}{\lambda} & =Z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=(3)^2 R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=8 R \\
\Rightarrow \text { Wavelength, } \lambda & =\frac{1}{8 R}=\frac{1}{8 \times 1.097 \times 10^7}=0.114 \times 10^{-7} \\
& =11.4 \mathrm{~nm}
\end{aligned}
\)
A hydrogen like atom of atomic number \(Z\) is in an excited state of quantum number \(2 n\). It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state \(n\), a photon of energy 40.8 eV is emitted. Find \(n, Z\) and the ground state energy (in eV ) for this atom. Also, calculate the minimum energy (in eV ) that can be emitted by this atom during de-excitation. (Take, ground state energy of hydrogen atom \(=13.6 \mathrm{eV}\) )
(b) Given, \(E_{2 n}-E_1=204 \mathrm{eV}\)
Using the relation \(\Delta E=-13.6 Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\), we get
\(
(13.6) Z^2\left(1-\frac{1}{4 n^2}\right)=204 \dots(i)
\)
\(
\begin{aligned}
&\text { Similarly, } \quad E_{2 n}-E_n=40.8 \mathrm{eV}\\
&\therefore 13.6 Z^2\left(\frac{1}{n^2}-\frac{1}{4 n^2}\right)=40.8 \dots(ii)
\end{aligned}
\)
Solving Eqs. (i) and (ii), we get
\(
n=2 \text { and } Z=4
\)
We know that, in ground state \((n=1), E=-13.6 Z^2 \mathrm{eV}\)
So,
\(
\begin{aligned}
E_1 & =(-13.6) Z^2 \mathrm{eV} \\
& =(-13.6)(4)^2 \mathrm{eV}=-217.6 \mathrm{eV}
\end{aligned}
\)
During de-excitation, minimum energy emitted is
\(
\begin{aligned}
E_{\min } & =E_{2 n}-E_{2 n-1}=E_4-E_3 \\
& =\frac{-217.6}{4^2}-\left(\frac{-217.6}{3^2}\right) \\
& =10.58 \mathrm{eV}
\end{aligned}
\)
A doubly ionised lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in \(\mathrm{Li}^{2+}\) from the first to the third Bohr orbit. The ionisation energy of the hydrogen atom is 13.6 eV.
(d) We know, \(E_n=-\frac{Z^2}{n^2}(13.6 \mathrm{eV})\)
By putting \(Z=3\) (given), we have
\(
E_n=-\frac{122.4}{n^2} \mathrm{eV}
\)
As, excitation is from 1st to 3rd Bohr orbit, i.e. \(n_1=1\) to \(n_2=3\).
\(
\Rightarrow \quad E_1=-\frac{122.4}{(1)^2}=-122.4 \mathrm{eV}
\)
and
\(
E_3=-\frac{122.4}{(3)^2}=-13.6 \mathrm{eV}
\)
\(
\therefore \quad \Delta E=E_3-E_1=-13.6-(-122.4)=108.8 \mathrm{eV}
\)
The corresponding wavelength is
\(
\lambda=\frac{12375}{\Delta E(\text { in eV })}=\frac{12375}{108.8}=113.74 Å
\)
Monochromatic radiation of wavelength \(\lambda\) is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of ten different wavelengths. Find the value of \(\lambda\).
(a) From \(n\)th energy state a total of \(\frac{n(n-1)}{2}\) lines are emitted.
\(
\therefore \quad \frac{n(n-1)}{2}=10 \text { or } n=5
\)
Further, \(\quad E_n=\frac{E_1}{n^2}\)
\(
\therefore \quad E_5=\frac{-13.6}{(5)^2}=-0.54 \mathrm{eV}
\)
The energy needed to take a hydrogen atom from its ground state to \(n=5\) is,
\(
\begin{aligned}
\Delta E=E_5-E_1 & =-0.54-(-13.6) \\
& =13.06 \mathrm{eV}
\end{aligned}
\)
The corresponding wavelength of light,
\(
\lambda=\frac{12375}{\Delta E} \AA=\frac{12375}{13.06}=947.5 Å
\)
Ionisation potential of hydrogen atom is 13.6 V. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy is 12.1 eV. According to Bohr’s theory, find the possible spectral lines emitted by hydrogen.
(c) Ionisation energy corresponding to ionisation potential
\(
=-13.6 \mathrm{eV}
\)
Energy of incident photon \(=12.1 \mathrm{eV}\)
So, the energy of electron in excited state
\(
=-13.6+12.1=-1.5 \mathrm{eV}
\)
As, \(\quad E_n=-\frac{13.6}{n^2} \mathrm{eV}\)
\(
\Rightarrow \quad-1.5=\frac{-13.6}{n^2} \Rightarrow n^2=\frac{-13.6}{-1.5}=9 \Rightarrow n=3
\)
i.e. the energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd, 3rd to 1st and 2nd to 1st. Thus, three spectral lines are emitted.
Kinetic energy of electron in \(n\)th orbit is given by
(d) The kinetic energy (KE) of an electron in the \(n\)th orbit is given by the formula:
\(
K E=\frac{2 \pi^2 m e^4}{n^2 h^2}
\)
where:
\(m\) is the mass of the electron,
\(e\) is the charge of the electron,
\(n\) is the principal quantum number (orbit number),
\(h\) is Planck’s constant.
The Rydberg constant \(R\) is defined as:
\(
R=\frac{2 \pi^2 m e^4}{c h^3}
\)
where \(c\) is the speed of light.
Rearrange the kinetic energy formula
We want to express the kinetic energy in terms of \(R, h[latex], and [latex]c\). To do this, we need to manipulate the original kinetic energy formula.
Modify the kinetic energy formula
We can multiply and divide the original kinetic energy formula by \(h[latex] and [latex]c\) :
\(
K E=\frac{2 \pi^2 m e^4}{n^2 h^2} \cdot \frac{h}{h} \cdot \frac{c}{c}
\)
This gives us:
\(
K E=\frac{2 \pi^2 m e^4 h c}{n^2 h^3}
\)
Substitute for \(R\)
Now, we can substitute \(R\) into the equation:
\(
K E=\frac{R h c}{n^2}
\)
Potential energy \(\left(\mathrm{PE}_n\right)\) and kinetic energy \(\left(\mathrm{KE}_n\right)\) of electron in \(n\)th orbit are related as
(b) The kinetic energy (KE) of an electron in the nth orbit is given by the formula:
\(
K E_n=\frac{Z e^2}{8 \pi \epsilon_0 r_n}
\)
The potential energy (PE) of the electron in the \(n\)th orbit is given by:
\(
P E_n=-\frac{Z e^2}{4 \pi \epsilon_0 r_n}
\)
Write the total energy ( E ) of the electron
The total energy (E) of the electron in the nth orbit is the sum of its kinetic and potential energy:
\(
E_n=K E_n+P E_n
\)
Substituting the expressions for KE and PE, we get:
\(
E_n=\frac{Z e^2}{8 \pi \epsilon_0 r_n}-\frac{Z e^2}{4 \pi \epsilon_0 r_n}
\)
Simplify the total energy expression
To combine the terms, we need a common denominator:
\(
E_n=\frac{Z e^2}{8 \pi \epsilon_0 r_n}-\frac{2 Z e^2}{8 \pi \epsilon_0 r_n}
\)
This simplifies to:
\(
E_n=-\frac{Z e^2}{8 \pi \epsilon_0 r_n}
\)
Relate PE and KE
From the expressions for KE and PE, we can derive the relationship between them:
We have:
\(
P E_n=-\frac{Z e^2}{4 \pi \epsilon_0 r_n}
\)
and
\(
K E_n=\frac{Z e^2}{8 \pi \epsilon_0 r_n}
\)
Now, we can express PE in terms of KE:
\(
P E_n=-2 \times K E_n
\)
In the \(n\)th orbit, the energy of an electron, \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\) for hydrogen atom. The energy required to take the electron from first orbit to second orbit will be
(a) For \(n=2\), using, \(E_n=\frac{-13.6 \mathrm{eV}}{n^2}, E_2=-\frac{13.6}{(2)^2}=-3.4 \mathrm{eV}\)
For \(n=1, \quad E_1=-13.6 \mathrm{eV}\)
\(
\therefore \quad E_{1 \rightarrow 2}=E_2-E_1=-3.4-(-13.6)=10.2 \mathrm{eV}
\)
Ionisation potential (IP) and ionisation energy (IE) are related as
(b) Ionization Energy (IE): This is the amount of energy required to remove an electron from an atom in its gaseous state.
Ionization Potential (IP): This is the potential difference (voltage) required to remove an electron from an atom.
Recall the Relationship
We know from electrostatics that energy can be expressed in terms of charge and potential:
Energy \(=\) Charge \(\times\) Potential
Apply the Definitions
In the context of ionization:
The energy required to remove an electron (IE) can be expressed as:
\(
I E=e \times I P
\)
where \(e\) is the charge of the electron.
Rearranging the Equation
From the equation \(I E=e \times I P\), we can rearrange it to express IP in terms of IE:
\(
I P=\frac{I E}{e}
\)
Conclusion
Thus, the relationship between ionization potential and ionization energy is:
\(
\begin{aligned}
& \text { Ionization Potential (IP) } \\
& =\frac{\text { Ionization Energy (IE) }}{e}
\end{aligned}
\)
In hydrogen atom, if the difference in the energy of the electron in \(n=2\) and \(n=3\) orbits is \(E\), the ionisation energy of hydrogen atom is
\(
\begin{aligned}
&\begin{aligned}
& \text { (b) } E_3-E_2=E \text { or } E=-13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=E_1\left(\frac{1}{9}-\frac{1}{4}\right) \\
& \Rightarrow \quad \quad E=\frac{E_1}{9}-\frac{E_1}{4} \quad \text { or } E_1=-7.2 E
\end{aligned}\\
&\therefore \text { Ionisation energy of hydrogen atom is } 7.2 E \text {. }
\end{aligned}
\)
Atomic hydrogen is excited from the ground state to the \(n\)th state. The number of lines in the emission spectrum will be
\(
\text { (b) Maximum no of spectral lines }=\frac{n(n-1)}{2}
\)
In which of the following transitions will the wavelength be minimum?
(d) The wavelength will be minimum for which \(\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) is maximum. Hence, option (d) is correct.
Which one of the series of hydrogen spectrum is in the visible region?
(b) Balmer series
Explanation:
Balmer series: This series of hydrogen spectral lines falls within the visible region of the electromagnetic spectrum. The wavelengths in the Balmer series correspond to transitions of electrons from higher energy levels to the second energy level ( \({n}=2\) ).
Lyman series: This series is in the ultraviolet region. It corresponds to transitions of electrons to the lowest energy level ( \({n}=1\) ).
Paschen series and Brackett series: These series, along with the Pfund series, are in the infrared region. They correspond to transitions of electrons to higher energy levels than the second level ( \(n=3\) for Paschen, \({n}=4\) for Brackett).
Key point: The Balmer series is the only hydrogen spectral series visible to the human eye.
The ratio of the largest to shortest wavelengths in Balmer series of hydrogen spectrum is
(c)
\(
\begin{gathered}
\lambda \propto \frac{1}{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)} \\
\frac{\lambda_{\text {largest }}}{\lambda_{\text {shortest }}}=\frac{\left(\frac{1}{4}-\frac{1}{\infty}\right)}{\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{(1 / 4)}{(5 / 36)}=\frac{9}{5}
\end{gathered}
\)
Wavelength corresponding to series limit is
(c) The wavelength corresponding to the series limit is given by \(\frac{1}{\lambda}=\frac{R}{n^2}\), where R is the Rydberg constant and n is the principal quantum number of the lower energy level the electron transitions to. For the series limit, \(n\) is typically 1 for the Lyman series, 2 for the Balmer series, etc.
For the Bohr’s first orbit of circumference \(2 \pi r\), the de-Broglie wavelength of revolving electron will be
(a) According to Bohr’s first postulate,
\(
\begin{aligned}
m v r & =\frac{n h}{2 \pi} \\
\therefore \quad 2 \pi r & =n\left(\frac{h}{m v}\right)=n \lambda
\end{aligned}
\)
For \(\quad n=1, \lambda=2 \pi r\)
Which of the following is incorrect regarding limitations of Bohr’s model?
(c) The incorrect statement regarding limitations of Bohr’s model is, “It’s assumption regarding stationary orbits supports Heisenberg uncertainty principle” because the Bohr model’s assumption of fixed, well-defined orbits directly contradicts the Heisenberg uncertainty principle, which states that knowing the position and momentum of an electron with precision is impossible.
Explanation:
Bohr model and stationary orbits:
Bohr’s model depicts electrons as orbiting the nucleus in fixed, circular paths with specific energy levels. This concept of stationary orbits is a key feature of the model.
Heisenberg uncertainty principle:
This principle states that you cannot simultaneously determine both the position and momentum of a particle with perfect accuracy. The more precisely you measure one, the less accurate your measurement of the other will be.
\(A 12.5 \mathrm{eV}\) electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
(a) Energy of electron beam, \(E=12.5 \mathrm{eV}\)
\(
\text { Planck’s constant, } h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}
\)
So, using the relation, \(E=\frac{h c}{\lambda}\), we get
\(
\begin{aligned}
\lambda=\frac{h c}{E} & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{12.5 \times 1.6 \times 10^{-19}} \\
& =0.993 \times 10^{-7}=993 \times 10^{-10} \mathrm{~m} \\
& =993 Å
\end{aligned}
\)
This wavelength falls in the range of Lyman series ( \(912 Å\) to \(1216 Å\) ), thus we conclude that Lyman series is emitted.
In \(H\)-atom, a transition takes place from \(n=3\) to \(n=2\) orbit. Calculate the wavelength of the emitted photon and will the photon be visible? To which spectral series will this photon belong? (Take, \(R=1.097 \times 10^7 \mathrm{~m}^{-1}\) )
(b) The wavelength of the emitted photon is given by
\(
\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
When the transition takes place from \(n=3\) to \(n=2\), then
\(
\begin{aligned}
\frac{1}{\lambda} & =\left(1.097 \times 10^7\right)\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& =1.097 \times 10^7 \times \frac{5}{36} \\
\therefore \quad \lambda & =\frac{36}{1.097 \times 10^7 \times 5}=6.563 \times 10^{-7} \mathrm{~m} \\
& =6563 Å
\end{aligned}
\)
\(\lambda\) falls in the visible (red) part of the spectrum, hence the photon will be visible. This photon is the first member of the Balmer series.
What is the shortest wavelength present in the Paschen series of spectral lines?
(b) For the shortest wavelength of Paschen series of spectral lines, \(n_1=3, n_2=\infty\). If \(\lambda_{\text {min }}\) be the minimum wavelength in this series of spectral lines, then
\(
\begin{aligned}
\frac{1}{\lambda_{\min }} & =R\left(\frac{1}{(3)^2}-\frac{1}{(\infty)^2}\right)=\frac{R}{9} \\
\lambda_{\min } & =\frac{9}{R}=\frac{9}{1.097 \times 10^7} \quad\left(\because R=1.097 \times 10^7 \mathrm{~m}^{-1}\right) \\
& =8204.2 \times 10^{-10} \mathrm{~m}=8204.2 Å
\end{aligned}
\)
You cannot copy content of this page