When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are V 1

Question

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are V 1 ​ and V 2 ​ respectively. The ratio V 1 ​ : V 2 ​ is

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{ma x} = E - Φ$ , where $E$ is the energy of the incident photon and $Φ$ is the work function of the metal.

For the first case, the incident energy is $E_{1} = 2Φ$ .

$K_{1} = 2Φ - Φ = Φ$

For the second case, the incident energy is $E_{2} = 3Φ$ .

$K_{2} = 3Φ - Φ = 2Φ$

The maximum kinetic energy is related to the maximum velocity by $K = \frac{1}{2} m V^{2}$ .

Taking the ratio of the kinetic energies in the two cases:

$\frac{K _{1}}{K _{2}} = \frac{\frac{1}{2} m V _{1}^{2}}{\frac{1}{2} m V _{2}^{2}} = \frac{Φ}{2Φ}$

$\frac{V _{1}^{2}}{V _{2}^{2}} = \frac{1}{2}$

Taking the square root of both sides:

$\frac{V _{1}}{V _{2}} = \frac{1}{2}$

The ratio $V_{1} : V_{2}$ is $1 : 2$ .

Answer: $1 : 2$

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