Given below are two statements: 

Assertion (A):	A photon has no rest mass, yet it carries definite momentum.
Reason (R):	Momentum of a photon is due to its energy and hence its equivalent mass.

 

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	Both (A) and (R) are False.

When ultraviolet rays strike a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

1. Infrared rays

2. X-rays

3. Radio wave

4. Lightwave

If the K.E. of an electron and a photon is the same, then the relation between their de-Broglie wavelength will be:

1. ${\lambda }_{ph}$ $<$ ${\lambda }_e$

2. ${\lambda }_{ph}$ $=$ ${\lambda }_e$

3. ${\lambda }_{ph}$ $>$ ${\lambda }_e$

4. ${\lambda }_{ph}$ $=2$ ${\lambda }_e$

A photosensitive metallic surface has a work function of hν₀. If photons of energy 2hν₀ fall on this surface, the electrons come out with a maximum velocity of 4 × 10⁶ m/s. When the photon energy is increased to 5hν₀, then the maximum velocity of photoelectrons will be:

1. 2 ×10⁷ m/s

2. 2 × 10⁶ m/s

3. 8 × 10⁵ m/s

4. 8 × 10⁶ m/s

The work function of a metal surface is φ = 1.5 eV. If a light of wavelength 5000 Å falls on it, then the maximum K.E. of the ejected electron will be:

The work functions for metals A, B, and C are respectively 1.92 eV, 2.0 eV, and 5 eV. According to Einstein's equation, the metals that will emit photoelectrons for a radiation of wavelength 4100 Å is/are:
1. None
2. A only
3. A and B only
4. All the three metals

A light of wavelength \(\lambda \) is incident on the metal surface and the ejected fastest electron has speed \(v.\) If the wavelength is changed to \(\frac{3\lambda}{4},\) then the speed of the fastest emitted electron will be:

What did Einstein prove by the photo-electric effect?

1. E = h\(\nu\)

2. $KE=\frac{1}{2}m{v}^2$

3. $E=m{c}^2$

4. $E=\frac{-Rh{c}^2}{n^2}$