A photosensitive metallic surface has a work function of \(h\nu_0\). If photons of energy \(2h\nu_0\) fall on this surface, the electrons come out with a maximum velocity of \(4\times10^{6}\) m/s. When the photon energy is increased to \(5h\nu_0\), then the maximum velocity of photoelectrons will be:
1. \(2\times 10^7~\text{m/s}\)
2. \(2\times 10^6~\text{m/s}\)
3. \(8\times 10^5~\text{m/s}\)
4. \(8\times 10^6~\text{m/s}\)

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

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 \(4100~\mathring{A}\) 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. \(K.E = \frac{1}{2}mv^2\)
3. \(E= mc^2\)
4. \(E = \frac{-Rhc^2}{n^2}\)

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. \(5~\text{V}\)

Consider a beam of electrons (each electron with energy \(E_0\)) incident on a metal surface kept in an evacuated chamber. Then:

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. \(102\times10^{-2}~\text{nm}\)