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

1.	no electrons will be emitted as only photons can emit electrons.
2.	electrons can be emitted but all with energy, \(E_0\)${\mathrm{}}_{}$.
3.	electrons can be emitted with any energy, with a maximum of \(\mathrm{E}_0-\phi\) (\(\phi\) is the work function).
4.	electrons can be emitted with any energy, with a maximum \(E_0\).

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

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

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. \(102\times10^{-2}~\text{nm}\)

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

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