The maximum kinetic energies of photoelectrons emitted are K 1 and K 2 when light of wavelength λ 1 and λ 2 respectively are incident on a metallic surface. If λ 1 = 3 λ 2 then

Question

The maximum kinetic energies of photoelectrons emitted are K 1 ​ and K 2 ​ when light of wavelength λ 1 ​ and λ 2 ​ respectively are incident on a metallic surface. If λ 1 ​ = 3 λ 2 ​ then

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K = \frac{h c}{λ} - Φ$ , where $Φ$ is the work function of the metal.

For wavelength $λ_{1}$ :

$K_{1} = \frac{h c}{λ _{1}} - Φ$

For wavelength $λ_{2}$ :

$K_{2} = \frac{h c}{λ _{2}} - Φ$

Given $λ_{1} = 3 λ_{2}$ , substituting this in the first equation gives:

$K_{1} = \frac{h c}{3 λ _{2}} - Φ$

Multiplying the equation by $3$ yields:

$3 K_{1} = \frac{h c}{λ _{2}} - 3Φ$

From the second equation, substituting $\frac{h c}{λ _{2}} = K_{2} + Φ$ into the above relation:

$3 K_{1} = (K_{2} + Φ) - 3Φ$

$3 K_{1} = K_{2} - 2Φ$

Since the work function $Φ > 0$ , it follows that:

$3 K_{1} < K_{2}$

$K_{1} < \frac{K _{2}}{3}$

Answer: $K_{1} < \frac{K _{2}}{3}$

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