JEE Physics Test Series Quiz-4

An electron, a doubly ionized helium ion \(\left(\mathrm{He}^{++}\right)\)and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths \(\lambda_{\mathrm{e}}, \lambda_{\mathrm{He}++}\) and \(\lambda_{\mathrm{p}}\) is:

A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is \(1.878 \times 10^{-4}\). The mass of the particle is close to:

A particle moving with kinetic energy \(\mathrm{E}\) has de Broglie wavelength \(\lambda\). If energy \(\Delta \mathrm{E}\) is added to its energy, the wavelength become \(\frac{\lambda}{2}\). Value of \(\Delta \mathrm{E}\), is:

A particle ‘ \(\mathrm{P}\) ‘ is formed due to a completely inelastic collision of particles ‘ \(x\) ‘ and ‘ \(y\) ‘ having de-Broglie wavelengths ‘ \(\gamma_x\) ‘ and ‘ \(\gamma_y\) ‘ respectively. If \(x\) and \(y\) were moving in opposite directions, then the de-Broglie wavelength of ‘ \(P\) ‘ is:

A particle of mass \(M\) at rest decays into two particles of masses \(m_1\) and \(m_2\), having non-zero velocities. The ratio of the de Broglie wavelengths of the particles, \(\lambda_1 / \lambda_2\), is

Two sources of light emit \(\mathrm{X}\)-rays of wavelength \(1 \mathrm{~nm}\) and visible light of wavelength \(500 \mathrm{~nm}\), respectively. Both the sources emit light of the same power \(200 \mathrm{~W}\). The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is:

The stopping potential \(\mathrm{V}_0\) (in volt) as a function of frequency (v) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given : Planck’s constant \((h)=6.63 \times 10^{-34} \mathrm{Js}\), electron charge \(e=1.6 \times 10^{-19} \mathrm{C}\) )

A metal plate of area \(1 \times 10^{-4} \mathrm{~m}^2\) is illuminated by a radiation of intensity \(16 \mathrm{~mW} / \mathrm{m}^2\). The work function of the metal is \(5 \mathrm{eV}\). The energy of the incident photons is \(10 \mathrm{eV}\) and only \(10 \%\) of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be:
\(
\left[1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right]
\)

A photoelectric surface is illuminated successively by monochromatic light of wavelengths \(\lambda\) and \(\frac{\lambda}{2}\). If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

Match List – I (Fundamental Experiment) with List – II (its conclusion) and select the correct option from the choice given below the list:

\(
\begin{array}{|l|l|}
\hline {\text { List-I }} & {\text { List-II }} \\
\hline \begin{array}{l}
\text { A. Franck-Hertz } \\
\text { Experiment }
\end{array} & \text { (i) Particle nature of light } \\
\hline \begin{array}{l}
\text { B. Photo-electric } \\
\text { experiment }
\end{array} & \text { (ii) Discrete energy levels of atom } \\
\hline \begin{array}{l}
\text { C. Davison-Germer } \\
\text { experiment }
\end{array} & \text { (iii) Wave nature of electron } \\
\hline & \text { (iv) Structure of atom } \\
\hline
\end{array}
\)

The anode voltage of a photocell is kept fixed. The wavelength \(\lambda\) of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

A pulse of light of duration \(100 \mathrm{~ns}\) is absorbed completely by a small object initially at rest. Power of the pulse is 30 \(\mathrm{mW}\) and the speed of light is \(3 \times 10^8 \mathrm{~ms}^{-1}\). The final momentum of the object is

\(K_\alpha\) wavelength emitted by an atom of atomic number \(Z=11\) is \(\lambda\). Find the atomic number for an atom that emits \(K_\alpha\) radiation with wavelength \(4 \lambda\).

In a photoelectric experiment, anode potential is plotted against plate current.

The potential difference applied to an X-ray tube is \(5 \mathrm{kV}\) and the current through it is \(3.2 \mathrm{~mA}\). Then the number of electrons striking the target per second is

The work function of a substance is \(4.0 \mathrm{eV}\). The longest wavelength of light that can cause photoelectron emission from this substance is approximately

The X-ray beam coming from an X-ray tube will be

The graph which depicts the results of Rutherford’s gold foil experiment with
\(\alpha\)-particles is:
\(\theta\) : Scattering angle
\(Y\): Number of scattered \(\alpha\)-particles detected (Plots are schematic and not to scale)

In a hydrogen atom, the electron makes a transition from \((n+1)^{\text {th }}\) level to the \(n^{\text {th }}\) level. If \(n \gg 1\), the frequency of radiation emitted is proportional to:

The time period of revolution of electron in its ground state orbit in a hydrogen atom is \(1.6 \times 10^{-16} \mathrm{~s}\). The frequency of revolution of the electron in its first excited state (in \(s^{-1}\) ) is:

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, \(\lambda_1 / \lambda_2\), of the photons emitted in this process is:

Radiation coming from transitions \(n=2\) to \(n=1\) of hydrogen atoms fall on \(\mathrm{He}^{+}\)ions in \(n=1\) and \(n=2\) states. The possible transition of helium ions as they absorb energy from the radiation is :

If the series limit frequency of the Lyman series is \(\nu_1\), then the series limit frequency of the P-fund series is:

According to Bohr’s theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the \(\mathrm{n}^{\text {th }}\) orbit is proportional to : ( \(\mathrm{n}=\) principal quantum number)

The binding energy of the electron in a hydrogen atom is \(13.6 \mathrm{eV}\), the energy required to remove the electron from the first excited state of \(\mathrm{Li}^{++}\)is:

The largest wavelength in the ultraviolet region of the hydrogen spectrum is \(122 \mathrm{~nm}\). The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

If the atom \(100 \mathrm{Fm}^{257}\) follows the Bohr model and the radius of \({ }_{100} \mathrm{Fm}^{257}\) is \(n\) times the Bohr radius, then find \(n\).

The transition from the state \(n=4\) to \(n=3\) in a hydrogenlike atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?

In hydrogen spectrum the wavelength of \(\mathrm{H}_\alpha\) line is 656 \(\mathrm{nm}\), whereas in the spectrum of a distant galaxy, \(\mathrm{H}_\alpha\) line wavelength is \(706 \mathrm{~nm}\). The estimated speed of the galaxy with respect to earth is,

As per Bohr model, the minimum energy (in \(\mathrm{eV}\) ) required to remove an electron from the ground state of doubly ionized \(\mathrm{Li}\) atom \((Z=3)\) is

Consider the spectral line resulting from the transition \(n=2 \rightarrow n=1\) in the atoms and ions given below. The shortest wavelength is produced by

The radius \(R\) of a nucleus of mass number \(A\) can be estimated by the formula \(R=\left(1.3 \times 10^{-15}\right) A^{1 / 3} \mathrm{~m}\). It follows that the mass density of a nucleus is of the order of : \(\left(M_{\text {prot }} \cong M_{\text {neut }} \simeq 1.67 \times 10^{-27} \mathrm{~kg}\right)\)

The ratio of the mass densities of nuclei of \({ }^{40} \mathrm{Ca}\) and \({ }^{16} \mathrm{O}\) is close to :

An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of \(8: 27\). The ratio of the radii of the nuclei (assumed to be spherical) is:

For uranium nucleus how does its mass vary with volume?

Order of magnitude of density of uranium nucleus is, \(\left[m_p=1.67 \times 10^{-27} \mathrm{~kg}\right.\) ]

Find the Binding energy per neucleon for \({ }_{50}^{120} \mathrm{Sn}\). Mass of proton \(m_p=1.00783 \mathrm{U}\), mass of neutron \(m_n=1.00867 \mathrm{U}\) and mass of tin nucleus \(m_{\mathrm{Sn}}=119.902199 \mathrm{U}\). (take \(1 \mathrm{U}=931 \mathrm{MeV}\) )

In a reactor, \(2 \mathrm{~kg}\) of \({ }_{92} \mathrm{U}^{235}\) fuel is fully used up in 30 days. The energy released per fission is \(200 \mathrm{MeV}\). Given that the Avogadro number, \(\mathrm{N}=6.023 \times 10^{26}\) per kilo mole and \(1 \mathrm{eV}\) \(=1.6 \times 10^{-19} \mathrm{~J}\). The power output of the reactor is close to:

Imagine that a reactor converts all given mass into energy and that it operates at a power level of \(10^{\circ}\) watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\) )

When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as : \({ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3 x+\mathrm{Q}\) (energy) where three particles named \(x\) are produced and energy \(Q\) is released. What is the name of the particle \(x\) ?

If a star can convert all the He nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is [Mass of He nucleus is \(4.0026 \mathrm{amu}\) and mass of Oxygen nucleus is \(15.9994 \mathrm{~amu}\) ]

Fast neutrons can easily be slowed down by

During a nuclear fusion reaction

A radioactive nucleus decays by two different processes. The half-life for the first process is \(10 \mathrm{~s}\) and that for the second is \(100 \mathrm{~s}\). The effective half-life of the nucleus is close to :

In a radioactive material, fraction of active material remaining after time \(t\) is \(9 / 16\). The fraction that was remaining after \(t / 2\) is :

Two radioactive substances \(A\) and \(B\) have decay constants \(5 \lambda\) and \(\lambda\) respectively. At \(t=0\), a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become \(\left(\frac{1}{e}\right)^2\) will be:

In a radioactive decay chain, the initial nucleus is \({ }_{90}^{232} \mathrm{Th}\). At the end there are \(6 \alpha\)-particles and \(4 \beta\)-particles which are emitted. If the end nucleus is \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}, \mathrm{A}\) and \(\mathrm{Z}\) are given by:

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At \(t=0\) it was 1600 counts per second and \(t=8\) seconds it was 100 counts per second. The count rate observed, as counts per second, at \(\mathrm{t}=6\) seconds is close to:

Let \(N_\beta\) be the number of \(\beta\) particles emitted by 1 gram of \(\mathrm{Na}^{24}\) radioactive nuclei (half-life \(=15 \mathrm{hrs}\) ) in 7.5 hours, \(\mathrm{N}_\beta\) is close to (Avogadro number \(=6.023 \times 10^{23} / \mathrm{g}\). mole):