What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 $\stackrel{0}{A}$?

1. ${10}^3 m{s}^{-1}$

2. $1.2\times {10}^3 m{s}^{-1}$

3. $1.4\times {10}^3 m{s}^{-1}$

4. $2.8\times {10}^3 m{s}^{-1}$

An electron and photon have the same wavelength. If E is the energy of photon and  p is the momentum of the electron, then the magnitude of $\frac{E}{p}$ in SI unit is:

1. $3.33\times {10}^{-9}$

2. $3.0\times {10}^8$

3. $1.1\times {10}^{-19}$

4. $9\times {10}^{16}$

The difference between kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500 $\stackrel{0}{A}$ and 500 $\stackrel{0}{A}$ will be

1. 1.61 J

2. 2.47 J

3. 3.96 J

4. $3.9\times {10}^{-19} J$

When a point source of light is 1 m away from a photoelectric cell, the photoelectric current is found to be I mA. If the same source is placed at 4 m from the same photoelectric cells, the photoelectric current (in mA) will be:

1. $\frac{I}{16}$

2. $\frac{I}{4}$

3. $4\mathrm{I}$

4. $16 \mathrm{I}$
 

The work function of tungsten and sodium is 4.5 eV and 2.3 eV respectively. If the threshold wavelength $\lambda$ for sodium is 5460 $\stackrel{0}{A}$, the value of $\lambda$ for tungsten is:
 

1. 2791 $\stackrel{0}{A}$

2. 3260 $\stackrel{0}{A}$

3. 1925 $\stackrel{0}{A}$

.4 1000 $\stackrel{0}{A}$

A photon of energy E ejects a photoelectrons from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is, given by:

1. $\sqrt{\frac{2m (W_0-E)}{eB}}$

2. $\sqrt{\frac{2e (E-W_0)}{mB}}$

3. $\sqrt{\frac{2m (E-W_0)}{eB}}$

4. $\sqrt{\frac{2m{W}_0}{eB}}$

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photoelectrons is:

1. 1.07 V

2. 2.66 V

3. 3.7 V

4. 4.81 V

If the work function for a certain metal is $3.2\times {10}^{-19} J$ and it is illuminated with light of frequency $v=8\times {10}^{14} Hz$, the maximum kinetic energy of the photoelectron would be:

1. $2.1\times {10}^{-19} J$

2. $3.2\times {10}^{-19} J$

3. $5.3\times {10}^{-19} J$

4. $8.5\times {10}^{-19} J$

Ultraviolet light of wavelength 300 nm and intensity 1.0 $W{m}^{-2}$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 $c{m}^2$ of the surface is nearly

1. $9.61\times {10}^{14} s^{-1}$

2. $4.12\times {10}^{13} s^{-1}$

3. $1.51\times {10}^{12} s^{-1}$

4. $2.13\times {10}^{11} s^{-1}$

A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become:

1. four times the original value

2. twice the original value

3. 1/6th of the original value

4. unchanged