A photon of energy 3.4 eV is incident on a metal having a work function of 2 eV. The maximum K.E. of photo-electrons is equal to:

For photoelectric emission from certain metals, the cutoff frequency is \(\nu\). If radiation of frequency \(2\nu\)$\mathrm{}$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
(\(m\) is the electron mass)

When monochromatic photons of wavelength \(4000\) Å are incident on the metal plate of work function \(2.1\) eV, what will be the stopping potential for the photocurrent?

The correct graph between the maximum energy of a photoelectron \(\left(K_{max}\right)\) and the inverse of the wavelength \(\left(\frac{1}{\lambda}\right)\) of the incident radiation is given by the curve:

A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photoelectric current for this light is 3V_o. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V_o. The photoelectric effect's threshold wavelength for this surface is?

1. 6λ

2. 4λ

3. λ/4

4. λ/6

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 variation of the kinetic energy \((K)\) of photoelectrons as a function of the frequency \((f)\) of the incident radiation is best shown by:

1.		2.
3.		4.

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 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