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Radiations of two photons energies, twice and five times the work function of metal are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be
  • a)
    1 : 1
  • b)
    1 : 2
  • c)
    1 : 3
  • d)
    1 : 4
Correct answer is option 'B'. Can you explain this answer?
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Explanation:
When a photon strikes a metal surface it can transfer its energy to an electron in the metal, causing the electron to be ejected from the metal, this process is called the photoelectric effect.

The maximum kinetic energy of photoelectrons is given by the equation:

Kmax = hf - ϕ

where Kmax is the maximum kinetic energy of the photoelectrons, h is Planck's constant, f is the frequency of the incident radiation, and ϕ is the work function of the metal.

Case 1: Energy of photon is twice the work function of the metal

In this case, the maximum kinetic energy of the photoelectrons emitted is given by:

Kmax1 = 2hf - ϕ

Case 2: Energy of photon is five times the work function of the metal

In this case, the maximum kinetic energy of the photoelectrons emitted is given by:

Kmax2 = 5hf - ϕ

Ratio of maximum velocity

The maximum velocity of photoelectrons can be calculated using the formula:

vmax = √(2Kmax/m)

where vmax is the maximum velocity of the photoelectrons, Kmax is the maximum kinetic energy of the photoelectrons, and m is the mass of the electron.

Taking the ratio of the maximum velocities for the two cases we get:

vmax1/vmax2 = √[(2Kmax1/m)/(2Kmax2/m)]

vmax1/vmax2 = √(Kmax1/Kmax2)

Substituting the values of Kmax1 and Kmax2, we get:

vmax1/vmax2 = √[(2hf - ϕ)/(5hf - ϕ)]

vmax1/vmax2 = √[2/5]

vmax1/vmax2 = 1:√(5/2)

vmax1/vmax2 = 1:2.24

Therefore, the ratio of the maximum velocity of photoelectrons emitted in the two cases is 1:2.24, which is closest to option B, 1:2.
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