When photons of energy \(h\nu\) fall on an aluminium plate (of work function \(E_0\)), photoelectrons of maximum kinetic energy \(K\) are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:
1. \(K+ E_0\)
2. \(2K\)
3. \(K\)
4. \(K + h\nu\)

The momentum of a photon of energy 1 MeV in kg m/s, will be :

1. $0.33\times {10}^6$

2. $7\times {10}^{-24}$

3. ${10}^{-22}$

4. $5\times {10}^{-22}$

An electron is accelerated through a potential difference of 10,000 V. Its de-Broglie wavelength is, (nearly) : ($m_e= 9\times {10}^{-31}kg$)

1. 12.2 nm

2. $12.2\times {10}^{-13}m$

3. $12.2\times {10}^{-12}m$

4. $12.2\times {10}^{-14}m$

Light quanta with an energy of 4.9 eV eject photoelectrons from a light photosensitive surface with the work function $\left(\mathrm{\varphi }\right)$ = 4.5 eV. The maximum impulse that can be transmitted to the surface when each electron ejected, is

1.  3.45 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

2.  4.35 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

3.  5.35 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

4.  4.53 x ${10}^{-25}$ kg m ${\mathrm{s}}^{-1}$

When monochromatic light of wavelength $\mathrm{\lambda }$ illuminates a metal surface then stopping potential for photoelectric current is 3${\mathrm{V}}_0$. If wavelength changes to 2$\mathrm{\lambda }$ then stopping potential becomes $V_0$. Stopping potential in case wavelength is changed to 3$\mathrm{\lambda }$, would be:

1.  $\frac{{\mathrm{V}}_0}{2}$

2.  $\frac{{\mathrm{V}}_0}{6}$

3.  $\frac{{\mathrm{V}}_0}{4}$

4.  $\frac{{\mathrm{V}}_0}{3}$

The stopping potential for photoelectrons emitted from a surface illuminated by the light of wavelength 400 nm is 500 mV. When the incident wavelength is changed to a new value, the stopping potential is found to be 800 mV. New wavelength is about:

1.  365 nm

2.  250 nm

3.  640 nm

4.  340 nm

Assume that sodium produces monochromatic light of wavelength $5.89 x {10}^{-7}$ m. At what approximate rate would a 10-watt sodium-vapor light be emitting photons? Assume that the efficiency of the light bulb is about 30%.

1.  8.9 x ${10}^{18}$ photons/s

2.  3.0 x ${10}^{19}$ photons/s

3.  9.9 x ${10}^{19}$ photons/s 

4.  2.0 x ${10}^{20}$ photons/s

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