If the work function for a certain metal is $3.2\times {10}^{-19}$ joule and it is illuminated with light of frequency $8\times {10}^{14}$ Hz. The maximum kinetic energy of the photo-electrons would be $\left(\mathrm{h}=6.63\times {10}^{-34} \mathrm{Js}\right)$

(a) $2.1\times {10}^{-19} \mathrm{J}$              (b) $8.5\times {10}^{-19} \mathrm{J}$ 
(c) $5.3\times {10}^{-19} \mathrm{J}$              (d) $3.2\times {10}^{-19 }\mathrm{J}$

The stopping potential for photoelectrons:

The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be 
(1) 12.4 eV                     

(2) 6.2 eV

(3) 100 eV                     

(4) 200 eV

When the light source is kept 20 cm away from a photo cell, stopping potential 0.6 V is obtained. When source is kept 40 cm away, the stopping potential will be 
1. 0.3 V                  2. 0.6 V
3. 1.2 V                  4. 2.4 V

The minimum energy required to remove an electron is called 
(1) Stopping potential               

(2) Kinetic energy

(3) Work function                     

(4) None of these

Assuming photoemission to take place, the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times, is 
(1) 4                       

(2) $\frac{1}{4}$

(3) 2                       

(4) $\frac{1}{2}$

If the work function of a metal is $\text{'}\mathrm{\varphi }\text{'}$ and the frequency of the incident light is '$\nu$', there is no emission of photoelectron if
(1) $\mathrm{v}<\frac{\mathrm{\varphi }}{\mathrm{h}}$                   

(2) $\mathrm{v}=\frac{\mathrm{\varphi }}{\mathrm{h}}$   

(3) $\mathrm{v}>\frac{\mathrm{\varphi }}{\mathrm{h}}$                   

(4) $\mathrm{v}>=\frac{\mathrm{\varphi }}{\mathrm{h}}$   

Light of wavelength $\mathrm{\lambda }$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to $\mathrm{\lambda }\text{'}$ where 
(1) $\mathrm{\lambda }\text{'}=\frac{\mathrm{\lambda }}{2}$           

(2) $\mathrm{\lambda }\text{'}=2\mathrm{\lambda }$

(3) $\frac{\mathrm{\lambda }}{2}<\mathrm{\lambda }\text{'}<\mathrm{\lambda }$      

(4) $\mathrm{\lambda }\text{'}>\mathrm{\lambda }$

The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is

(1) 4125 Å                 

(2) 2062.5 Å

(3) 3000 Å                 

(4) 6000 Å