For the intensity \(I\) of a light of wavelength \(5000~\mathring{A}\) the photoelectron saturation current is \(0.40~\mu\text{A}\) and the stopping potential is \(1.36~\text{V},\) the work function of the metal is:
1. \(2.47~\text{eV}\)
2. \(1.36~\text{eV}\)
3. \(1.10~\text{eV}\)
4. \(0.43~\text{eV}\)

The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the surfaces of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B) 
(a) 1 : 1               (b) 1 : 2
(c) 1 : 3               (d) 1 : 4

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