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(A) According to Einstein's photoelectric equation: $eV_S = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For two different wavelengths

Vedclass · Accepted Answer

$+ 0.34$

Vedclass · Answer

(A) According to Einstein's photoelectric equation: $eV_S = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For two different wavelengths $\lambda_1 = 4000 \ \mathring{A}$ and $\lambda_2 = 3600 \ \mathring{A}$,the stopping potentials are $V_{S1}$ and $V_{S2}$ respectively.$eV_{S1} = \frac{hc}{\lambda_1} - \phi$ and $eV_{S2} = \frac{hc}{\lambda_2} - \phi$.Subtracting the two equations: $e(V_{S2} - V_{S1}) = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} ight)$.$\Delta V_S = \frac{hc}{e} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} ight)$.Using $hc = 12400 \ 	ext{eV} \cdot \mathring{A}$:$\Delta V_S = 12400 \left( \frac{1}{3600} - \frac{1}{4000} ight) = 12400 \left( \frac{4000 - 3600}{3600 	imes 4000} ight)$.$\Delta V_S = 12400 \left( \frac{400}{14400000} ight) = \frac{12400}{36000} = \frac{124}{360} \approx 0.344 \ 	ext{V}$.Since the wavelength decreases,the energy of incident photons increases,leading to an increase in stopping potential. Thus,the change is $+0.34 \ 	ext{V}$.

If the wavelength of incident light is decreased from $4000 \ \mathring{A}$ to $3600 \ \mathring{A}$,then the change in stopping potential will be ............. $V$.

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