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

An image of the sun is formed by a lens of focal length of 30 cm on the metal surface of a photoelectric cell and a photoelectric current I is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photoelectric current in this case is 
(1) $\frac{\mathrm{I}}{2}$                   

(2) I

(3) 2I                     

(4) 4I

A photon of $1.7\times {10}^{-13}$ Joules is absorbed by a material under special circumstances. The correct statement is:

1. Electrons of the atom of absorbed material will go the higher energy states

2. Electron and positron pair will be created

3. Only positron will be produced

4. Photoelectric effect will occur and electron will be produced

The maximum velocity of an electron emitted by light of wavelength $\mathrm{\lambda }$ incident on the surface of a metal of work function $\mathrm{\varphi }$ is 
(1) ${\left[\frac{2\left(\mathrm{hc}+\mathrm{\lambda \varphi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$       

(2) ${\left[\frac{2\left(\mathrm{hc}+\mathrm{\lambda \varphi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$

(3) ${\left[\frac{2\left(\mathrm{hc}-\mathrm{\lambda \varphi }\right)}{\mathrm{m\lambda }}\right]}^{1/2}$       

(4) ${\left[\frac{2\left(\mathrm{h\lambda }-\mathrm{\varphi }\right)}{\mathrm{m}}\right]}^{1/2}$

Where h = Planck's constant, m = mass of electron and c = speed of light.

In a photoemissive cell with executing wavelength $\mathrm{\lambda }$, the fastest electron has speed v. If the exciting wavelength is changed to 3$\mathrm{\lambda }$/4, the speed of the fastest emitted electron will be 
(a) $\mathrm{v}{\left(3/4\right)}^{1/2}$                              (b) $\mathrm{v}{\left(4/3\right)}^{1/2}$
(c) Less than $\mathrm{v}{\left(4/3\right)}^{1/2}$               (d) Greater than $\mathrm{v}{\left(4/3\right)}^{1/2}$

Photoelectric emission is observed from a metallic surface for frequencies ${\mathrm{v}}_1$ and ${\mathrm{v}}_2$ of the incident light rays $\left({\mathrm{v}}_1>{\mathrm{v}}_2\right)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1:k, then the threshold frequency of the metallic surface is

(1) $\frac{{\mathrm{v}}_1-{\mathrm{v}}_2}{\mathrm{k}-1}$          

(2) $\frac{{\mathrm{kv}}_1-{\mathrm{v}}_2}{\mathrm{k}-1}$

(3) $\frac{{\mathrm{kv}}_2-{\mathrm{v}}_1}{\mathrm{k}-1}$         

(4) $\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{\mathrm{k}}$

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27 $°\mathrm{C}$ and 127 $°\mathrm{C}$ respectively is
(1) $\frac{1}{2}$               

(2) $\sqrt{\frac{3}{8}}$

(3) $\sqrt{\frac{8}{3}}$             

(4) 1