In a photoelectric experiment, it was found that the stopping potential decreases from 2.5 V to V as the wavelength of the light is varied from 3$\mathrm{\lambda }$ to 4$\mathrm{\lambda }$. Calculate the Planck's constant in terms of V, $\mathrm{\lambda }$ and speed of light (c).

1.  $\frac{9\mathrm{\lambda eV}}{\mathrm{c}}$

2.  $\frac{6\mathrm{\lambda eV}}{\mathrm{c}}$

3.  $\frac{9\mathrm{\lambda eV}}{4\mathrm{c}}$

4.  $\frac{18\mathrm{\lambda eV}}{\mathrm{c}}$

The photoelectric effect is observed for two frequencies of 3 x ${10}^{14}$ Hz and 2 x ${10}^{14}$ Hz of incident radiation. If maximum kinetic energies are in ratio 2:1, then threshold frequency is
 

1.  ${10}^{14}$ Hz

2.  $\frac{3}{2} \times {10}^{14} \mathrm{Hz}$

3.  $\frac{4}{3}\times {10}^{14} \mathrm{Hz}$

4.  None of these

An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de-Broglie wavelength is $\lambda$. Then:

1.  E = 6.8 eV, $\mathrm{\lambda }$ = 6.6 x ${10}^{-10}$ m

2.  E = 3.4 eV, $\mathrm{\lambda }$ = 6.6 x ${10}^{-10}$ m

3.  E = 6.6 eV, $\mathrm{\lambda }$ = 6.6 x ${10}^{-11}$ m

4.  E = 6.8 eV, $\mathrm{\lambda }$ = 6.6 x ${10}^{-11}$ m

Radiation of wavelength $\frac{{\mathrm{\lambda }}_0}{3}$ (${\mathrm{\lambda }}_0$ being the threshold wavelength) is incident on a photosensitive sphere of radius R. The charge developed on the sphere when electrons cease to be emitted will be-

1.  $\frac{4{\mathrm{\pi \epsilon }}_0\mathrm{Rhc}}{{\mathrm{e\lambda }}_0}$

2.  $\frac{6{\mathrm{\pi \epsilon }}_0\mathrm{Rhc}}{{\mathrm{e\lambda }}_0}$

3.  $\frac{8{\mathrm{\pi \epsilon }}_0\mathrm{Rhc}}{{\mathrm{e\lambda }}_0}$

4.  $\frac{12{\mathrm{\pi \epsilon }}_0\mathrm{Rhc}}{{\mathrm{e\lambda }}_0}$

In an experiment on the photoelectric effect. the wavelength of the incident radiation is $\lambda$. The wavelength of the incident radiation is reduced to $\frac{1}{3}$rd of the initial value and the maximum kinetic energy of the photoelectron is observed to be n times the previous value. The threshold wavelength for the metal plate is :

1.  $\left(\frac{\mathrm{n} - 1}{\mathrm{n} - 3}\right)\mathrm{\lambda }$

2.  $\left(\frac{\mathrm{n}}{\mathrm{n} - 3}\right)\mathrm{\lambda }$

3.  $\frac{\left(\mathrm{n} + 1\right)\mathrm{\lambda }}{\left(\mathrm{n} - 3\right)}$

4.  $\frac{\mathrm{\lambda }}{\mathrm{n}}$

According to the Bohr model of the atom, an electron undergoes a transition from one orbit that is closer to the nucleus to another which is farther from the nucleus by absorbing a photon whose energy E depends on its frequency f as E = hf, where h is Planck's constant. The energy of 13.6 eV is needed to ionize a hydrogen atom by ejecting an electron from the lowest energy level. What is the longest wavelength of a photon that can eject the electron from the lowest energy level of the atom?

1.  40 nm

2.  60 nm

3.  80 nm

4.  90 nm

Photons of energy 7 eV are incident on two metals A and B with work functions 6 eV and 3 eV respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum energies are ${\lambda }_A$ and ${\lambda }_B$, respectively where $\frac{{\lambda }_A}{{\lambda }_B}$ is nearly-

1.  0.5

2.  1.4

3.  4.0

4.  2.0

A 160 watt light source is radiating light of wavelength 6200 Å uniformly in all directions. The photon flux at a distance of 1.8 m is of the order of (Planck's constant $6.63 \mathrm{x} {10}^{-34} \mathrm{J}-\mathrm{s}$) :

1. ${10}^2$  ${\mathrm{s}}^{-1}$

2. ${10}^{12}$  ${\mathrm{s}}^{-1}$

3. ${10}^{19}$  ${\mathrm{s}}^{-1}$

4. ${10}^{25}$  ${\mathrm{s}}^{-1}$

An electron in an electron microscope with initial velocity ${\mathrm{V}}_0\hat{i}$ enters a region of a strong transverse electric field ${\mathrm{E}}_0\hat{\mathrm{j}}$. The time taken for the change in its de-Broglie wavelength from the initial value of $\mathrm{\lambda }$ to $\mathrm{\lambda }/3$ is proportional to :

1.  ${\mathrm{E}}_0$

2.  $\frac{1}{{\mathrm{E}}_0}$

3.  $\frac{1}{\sqrt{{\mathrm{E}}_0}}$

4.  $\sqrt{{\mathrm{E}}_0}$

In the case of the photoelectric effect -

1. Since photons are absorbed as a single unit, there is no significant time delay in the emission of photoelectrons.

2. Einstein's analysis gives a critical frequency ${\nu }_0 = \frac{e\varphi }{h}$, where $\varphi$ is the work function and the light of this frequency ejects electrons with maximum kinetic energy.

3. Only a small fraction of the incident photons succeed in ejecting photoelectrons while most of them are absorbed by the system as a whole and generate thermal energy.

4. The maximum kinetic energy of the electrons is dependent on the intensity of radiation.