Single Choice

Threshold wavelength of a metal is $${\lambda}_{0}$$. The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength $$\lambda$$ is:

A$${\left[\dfrac{h \lambda {\lambda}_{0}}{2mc}\right]}^{\dfrac{1}{2}}$$
B$${\left[\dfrac{h \left(\lambda - {\lambda}_{0}\right)}{2mc \lambda {\lambda}_{0}}\right]}^{{1}/{2}}$$
C$${\left[\dfrac{h \left({\lambda}_{0} - \lambda\right)}{2mc \lambda {\lambda}_{0}}\right]}^{{1}/{2}}$$
D$${\left[\dfrac{h \lambda {\lambda}_{0}}{2mc \left({\lambda}_{0} - \lambda\right)}\right]}^{{1}/{2}}$$
Correct Answer

Solution

$$\mathbf{Explanation:}$$

From $$\mathit{Einstein's\ photoelectric\ equation}$$ we get:

$$\mathbf{\cfrac{hc}{\lambda }=\cfrac{hc}{\lambda _{th}}+K.E}$$

where,

$$\lambda =$$Wavelength

$$\lambda _{th}=$$threshold wavelength

$$h=$$Planck's constant

$$c=$$Velocity of light

$$\because K.E=\cfrac{1}{2}mv^{2}=\cfrac{hc}{\lambda }-\cfrac{hc}{\lambda _{th}}=hc\left [ \cfrac{1}{\lambda }-\cfrac{1}{\lambda _{th}} \right ]$$

$$\mathit{De\ Broglie's\ wavelength}$$ attached to it,$$\lambda _{d}=\cfrac{h}{p}=\cfrac{h}{mv}$$

$$\therefore \lambda _{d}=\cfrac{h}{\sqrt{2K.E\times m}}=\cfrac{h}{\sqrt{2\times hc(\cfrac{1}{\lambda }-\cfrac{1}{\lambda _{th}})\times m}}$$

$$=\cfrac{h}{\sqrt{2\times hmc\left ( \cfrac{\lambda _{th}-\lambda }{\lambda \lambda _{th}} \right )}}$$

$$=[\cfrac{h^{2}\lambda \lambda _{th}}{2hmc(\lambda _{th}-\lambda )}]^{\frac{1}{2}}=[\cfrac{h\lambda \lambda _{th}}{2mc(\lambda _{th}-\lambda )}]^\frac{1}{2}$$

$$\mathbf{\therefore \lambda _{d}=[\cfrac{h\lambda \lambda _{th}}{2mc(\lambda _{th}-\lambda )}]}$$

where $$\lambda_{th}= \lambda_o$$

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D).}$$

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