When a metal surface is illuminated by light | vedclass

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Question

(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.For $\lambda_1 = 400\; nm =

Vedclass · Accepted Answer

$2hc 	imes 10^6\; J$

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(A) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.For $\lambda_1 = 400\; nm = 400 	imes 10^{-9}\; m$,velocity is $v$: $\frac{hc}{400 	imes 10^{-9}} = W_0 + \frac{1}{2}mv^2$ ... $(i)$For $\lambda_2 = 250\; nm = 250 	imes 10^{-9}\; m$,velocity is $2v$: $\frac{hc}{250 	imes 10^{-9}} = W_0 + \frac{1}{2}m(2v)^2 = W_0 + 2mv^2$ ... $(ii)$From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{400 	imes 10^{-9}} - W_0$. Substituting this into $(ii)$:$\frac{hc}{250 	imes 10^{-9}} = W_0 + 4(\frac{hc}{400 	imes 10^{-9}} - W_0)$$\frac{hc}{250 	imes 10^{-9}} = W_0 + \frac{hc}{100 	imes 10^{-9}} - 4W_0$$3W_0 = hc(\frac{1}{100 	imes 10^{-9}} - \frac{1}{250 	imes 10^{-9}}) = hc(\frac{2.5 - 1}{250 	imes 10^{-9}}) = hc(\frac{1.5}{250 	imes 10^{-9}})$$3W_0 = hc(0.006 	imes 10^9) = hc(6 	imes 10^6)$$W_0 = 2hc 	imes 10^6\; J$.

When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

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