In an experiment on the photoelectric effect, | vedclass

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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by: $(KE)_{\max} = hc \left( \frac{1}{\lambda} -

Vedclass · Accepted Answer

$\left( \frac{n - 1}{n - 3} \right) \lambda$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by: $(KE)_{\max} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$ where $\lambda_0$ is the threshold wavelength. For the initial case: $(KE)_1 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$  .........$(i)$ When the wavelength is reduced to $\frac{\lambda}{3}$,the new kinetic energy is $n$ times the initial value: $(KE)_2 = n(KE)_1 = hc \left( \frac{1}{\lambda/3} - \frac{1}{\lambda_0} \right) = hc \left( \frac{3}{\lambda} - \frac{1}{\lambda_0} \right)$  .........$(ii)$ Dividing equation $(ii)$ by equation $(i)$: $n = \frac{\frac{3}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{\lambda} - \frac{1}{\lambda_0}} = \frac{\frac{3\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}} = \frac{3\lambda_0 - \lambda}{\lambda_0 - \lambda}$ $n(\lambda_0 - \lambda) = 3\lambda_0 - \lambda$ $n\lambda_0 - n\lambda = 3\lambda_0 - \lambda$ $n\lambda_0 - 3\lambda_0 = n\lambda - \lambda$ $\lambda_0(n - 3) = \lambda(n - 1)$ $\lambda_0 = \left( \frac{n - 1}{n - 3} \right) \lambda$

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