If v_0 is the threshold frequency of a metal | vedclass

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Question

(A) According to the de Broglie wavelength equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$.Since kinetic energy $KE = \frac{1}{2}mv^2$,we have $v = \s

Vedclass · Accepted Answer

$\lambda \propto \frac{1}{\sqrt{v-v_0}}$

Vedclass · Answer

(A) According to the de Broglie wavelength equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$.Since kinetic energy $KE = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2 	imes KE}{m}}$.Substituting this into the wavelength equation: $\lambda = \frac{h}{\sqrt{2m 	imes KE}}$.From the photoelectric effect equation,$KE = hv - hv_0 = h(v - v_0)$.Substituting $KE$ into the wavelength expression: $\lambda = \frac{h}{\sqrt{2mh(v - v_0)}} = \sqrt{\frac{h^2}{2mh(v - v_0)}} = \sqrt{\frac{h}{2m(v - v_0)}}$.Therefore,$\lambda \propto \frac{1}{\sqrt{v - v_0}}$ or $\lambda \propto (v - v_0)^{-\frac{1}{2}}$.

If $v_0$ is the threshold frequency of a metal $X$,the correct relation between de Broglie wavelength $(\lambda)$ associated with the photoelectron and the frequency $(v)$ of the incident radiation is:

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