An electron with \(144~\text{eV}\) of kinetic energy has a de-Broglie wavelength that is very similar to?
1. \(102\times10^{-3}~\text{nm}\)
2. \(102\times10^{-4}~\text{nm}\)
3. \(102\times10^{-5}~\text{nm}\)
4. \(102\times10^{-2}~\text{nm}\)

Two radiations of photons energies \(1\) eV and \(2.5\) eV, successively illuminate a photosensitive metallic surface of work function \(0.5\) eV. The ratio of the maximum speeds of the emitted electrons is:
1. \(1:2\)
2. \(1:1\)
3. \(1:5\) 
4. \(1:4\)

If the momentum of an electron is changed by \(P,\) then the de-Broglie wavelength associated with it changes by \(0.5\%.\) The initial momentum of an electron will be:
1. \(400P\)
2. \(\frac{P}{100}\)
3. \(100P\)
4. \(200P\)

The threshold frequency for a photosensitive metal is \(3.3\times10^{14}~\text{Hz}\). If the light of frequency \(8.2\times10^{14}~\text{Hz}\) is incident on this metal, the cutoff voltage for the photoelectric emission will be:
1. \(1~\text{V}\)
2. \(2~\text{V}\)
3. \(3~\text{V}\)
4. \(5~\text{V}\)

When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively:

1. 2N and T

2. 2N and 2T

3. N and T

4. N and 2T

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. $E=\frac{-Rh{c}^2}{n^2}$

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:

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

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:

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