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An electromagnetic wave of wavelength $'\lambda '$ ....

An electromagnetic wave of wavelength $'\lambda '$ is incident on a photosensitive surface of negligible work function. If 'm' mass of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then:

(1) $\lambda  = \left( {\frac{{2m}}{{hc}}} \right){\lambda _d}^2$
(2) ${\lambda _d} = \left( {\frac{{2mc}}{h}} \right){\lambda ^2}$
(3) $\lambda  = \left( {\frac{{2mc}}{h}} \right){\lambda _d}^2$
(4) $\lambda  = \left( {\frac{{2h}}{{mc}}} \right){\lambda _d}^2$

Solution

de-Broglie wavelength $\lambda_d = \frac {h}{p}$

p can be obtained from photoelectric effect equation $h\nu  = \phi  + \frac{1}{2}m{v^2}$

Or $\frac{{hc}}{\lambda } = 0 + \frac{1}{2}\frac{{{p^2}}}{m}$

$\therefore p = \sqrt {\left( {\frac{{2mc}}{\lambda }} \right)h} $

Now, ${\lambda _d} = \frac{h}{p} = \frac{h}{{\sqrt {\left( {\frac{{2mc}}{\lambda }} \right)h} }} = \sqrt {\left( {\frac{h}{{2mc}}} \right)\lambda } $

$\therefore \lambda  = \left( {\frac{{2mc}}{h}} \right){\lambda _d}^2$

Answer: (3)

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