The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3 n , the maximum velocity of the ejected photoelectrons will be:

Question

Aicharya · Accepted Answer

According to Einstein's photoelectric effect,

$K E_{\max} = \frac{h c}{λ} - ϕ$

The incident energy of the photon is $E_{1} = \frac{1}{2} m v^{2} = h n - ϕ . . . (1)$

If incident frequency increased to 3n

$E_{2} = \frac{1}{2} m v_{2}^{2} = 3 h n - ϕ$

$= 3 (h n - ϕ) + 2 ϕ$

$E_{2} = 3 E_{1} + 2 ϕ$

$\Rightarrow \frac{1}{2} m v_{2}^{2} = \frac{3}{2} m v^{2} + 2 ϕ$

$\Rightarrow v_{2}^{2} = 3 v^{2} + \frac{4 ϕ}{m}$

$\Rightarrow v_{2} > v \sqrt{3}$

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on…

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