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

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

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 is accelerated from rest through a potential difference of \(V\) volt. If the de Broglie wavelength of an electron is $\mathrm{}$\(1.227\times10^{-2}~\text{nm}\). what will be its potential difference?
1. \(10^{2}~\text{V}\)
2. \(10^{3}~\text{V}\)
3. \(10^{4}~\text{V}\)
4. \(10^{5}~\text{V}\)

The de-Broglie wavelength of a particle accelerated by a 150 volt potential difference is 10^-10 m. What will its wavelength be if it is accelerated by a 600 volt potential difference?

1. 0.25 $\stackrel{o}{A}$

2. 0.5 $\stackrel{o}{A}$

3. 1.5 $\stackrel{o}{A}$

4. 2 $\stackrel{o}{A}$

In an experiment of the photoelectric effect, the wavelength of incident radiation is $\lambda$. The wavelength of incident radiation is reduced to $\frac{1}{3}$rd of initial value and the maximum kinetic energy of photoelectron is observed to be n times the previous value. What will be the threshold wavelength for the metal plate?
1. \(\frac{n-1}{n-3} \lambda \)
2. \(\frac{n}{n-3} \lambda \)
3. \(\frac{n-3}{n-1} \lambda \)
4. \(\frac{n+1}{n-3} \lambda\)

A source S₁ is producing, 10¹⁵ photons per sec of wavelength 5000 Å. Another source S₂ is producing 1.02×10¹⁵ photons per second of wavelength 5100 Å. Then the ratio of the power of S₂ to the power of S₁ is equal to: