When a photon of light collides with a metal surface, the number of electrons, (if any) coming out is:

1. only one

2. only two

3. infinite

4. depends upon factors

A point source of light is used in the photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential:

1. will increase

2. will decrease

3. will remain constant

4. will either increase or decrease

What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. ( Atomic mass of nitrogen = 14.0076 u)

1 0.01 nm

2 0.09 nm 

3 0.03 nm 

4 0.2 nm

In a photoelectric experiment, electrons are ejected from metals X and Y by the light of intensity I and frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If Y has a greater work function than X; which one of the following graphs best illustrates the expected results?

1. 

2. 

3. 

4.

An electron with the initial kinetic energy of 100 eV is accelerated through a potential difference of 50 V. Now the de-Broglie wavelength of electron becomes-

1. 1 $\stackrel{0}{A}$

2. $\sqrt{1.5}$ $\stackrel{0}{A}$

3. $\sqrt{3}$$\stackrel{0}{A}$

4. 12.27 $\stackrel{0}{A}$

The de-Broglie wavelength L associated with an elementary particle of linear momentum p is best represented by the graph:

1. 

2. 

3. 

4. 

An electron of mass m, when accelerated through a potential difference, has de-Broglie wavelength $\lambda$ .The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:

1. $\lambda \frac{m}{M}$

2. $\lambda \sqrt{\frac{m}{M}}$

3. $\lambda \frac{M}{m}$

4. $\lambda \sqrt{\frac{M}{m}}$

An electron (mass m) with an initial velocity v=$v_0\hat{i}$ is in an electric E=$E_o\hat{j}$. If ${\lambda }_0=h/m{v}_0$, its de-Broglie wavelength at time t is given by

1. ${\lambda }_0$

2. ${\lambda }_0\sqrt{1+\frac{e^2{E}_0^2{t}^2}{m^2{v}_0^2}}$

3. $\frac{{\lambda }_0}{\sqrt{1+{\displaystyle \frac{e^2{E^2}_0{t}^2}{m^2{v^2}_0}}}}$

4. $\frac{{\lambda }_0}{\left(1+{\displaystyle \frac{e^2{E^2}_0{t}^2}{m^2{v^2}_0}}\right)}$

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. $9\times {10}^{16}$