4 eV is the energy of the incident photon and the work function in 2eV. What is the stopping potential ?

(1) 2V                   

(2) 4V

(3) 6V                   

(4) $2\sqrt{2} \mathrm{V}$ 

The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is (taking h = $6\times {10}^{-34}$J-sec)

(a) 100                 (b) 1000
(c) $3\times {10}^{20}$           (d) $3\times {10}^{18}$

Light of frequency 4${\mathrm{v}}_0$ is incident on the metal of the threshold frequency ${\mathrm{v}}_0$. The maximum kinetic energy of the emitted photoelectrons is 

(1) $3{\mathrm{hv}}_0$          

(2) $2{\mathrm{hv}}_0$

(3) $\frac{3}{2}{\mathrm{hv}}_0$         

(4) $\frac{1}{2}{\mathrm{hv}}_0$

Two identical photo-cathodes receive light of frequencies ${\mathrm{f}}_1$ and ${\mathrm{f}}_2$. If the velocities of the photo electrons (of mass m) coming out are respectively ${\mathrm{v}}_1$ and ${\mathrm{v}}_2$, then 
(1) ${\mathrm{v}}_1-{\mathrm{v}}_2={\left[\frac{2\mathrm{h}}{\mathrm{m}}\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)\right]}^{1/2}$             

(2) ${\mathrm{v}}_1^2-{\mathrm{v}}_2^2=\frac{2\mathrm{h}}{\mathrm{m}}\left({\mathrm{f}}_1-{\mathrm{f}}_2\right)$

(3) ${\mathrm{v}}_1+{\mathrm{v}}_2={\left[\frac{2\mathrm{h}}{\mathrm{m}}\left({\mathrm{f}}_1+{\mathrm{f}}_2\right)\right]}^{1/2}$             

(4) ${\mathrm{v}}_1^2+{\mathrm{v}}_2^2=\frac{2\mathrm{h}}{\mathrm{m}}\left({\mathrm{f}}_1+{\mathrm{f}}_2\right)$

When radiation of wavelength $\mathrm{\lambda }$ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is

(a) $2\mathrm{\lambda }$                 (b) $4\mathrm{\lambda }$
(c) $6\mathrm{\lambda }$                 (d) $8\mathrm{\lambda }$

If the energy of the photon is increased by a factor of 4, then its momentum 

(1) Does not change

(2) Decreases by a factor of 4

(3) Increases by a factor of 4

(4) Decreases by a factor of 2

The work function for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein’s equation, the metals which will emit photo electrons for a radiation of wavelength 4100 Å is/are 
(1) None of these               

(2) A only

(3) A and B only                 

(4) All the three metals

The magnitude of saturation photoelectric current depends upon 
(1) Frequency             

(2) Intensity

(3) Work function       

(4) Stopping potential

The light rays having photons of energy \(1.8~\text{eV}\) are falling on a metal surface having a work function \(1.2~\text{eV}\). What is the stopping potential to be applied to stop the emitting electrons:
1. \(3~\text{eV}\) 
2. \(1.2~\text{eV}\)
3. \(0.6~\text{eV}\)
4. \(1.4~\text{eV}\)

A photon and an electron have equal energy E, then ${\mathrm{\lambda }}_{\mathrm{photon}}/{\mathrm{\lambda }}_{\mathrm{electron}}$ is proportional to

(1) $\sqrt{\mathrm{E}}$                     

(2) $1/\sqrt{\mathrm{E}}$

(3) 1/E                       

(4) Does not depend upon E