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Question

(A) Using Einstein's photoelectric equation,$h 
u = \phi_0 + KE_{	ext{max}}$.Since $KE_{	ext{max}} = eV_s$,we have $\frac{hc}{\lambda} = \phi_0 + e

Vedclass · Accepted Answer

$9 \lambda$

Vedclass · Answer

(A) Using Einstein's photoelectric equation,$h 
u = \phi_0 + KE_{	ext{max}}$.Since $KE_{	ext{max}} = eV_s$,we have $\frac{hc}{\lambda} = \phi_0 + e(4V_0)$ ....$(i)$Similarly,for wavelength $3\lambda$,we have $\frac{hc}{3\lambda} = \phi_0 + eV_0$ ....(ii)Multiply equation (ii) by $4$:$\frac{4hc}{3\lambda} = 4\phi_0 + 4eV_0$ ....(iii)Subtract equation $(i)$ from equation (iii):$\frac{4hc}{3\lambda} - \frac{hc}{\lambda} = (4\phi_0 + 4eV_0) - (\phi_0 + 4eV_0)$$\frac{hc}{3\lambda} = 3\phi_0$$\phi_0 = \frac{hc}{9\lambda}$Since the work function $\phi_0 = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength,we get $\lambda_0 = 9\lambda$.Thus,option $(A)$ is correct.

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