Question

An electromagnetic wave of wavelength $\lambda$is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Broglie wavelength $\lambda'$then:

• A. $\lambda = \frac{mc}{h}\lambda'^{2}$
• B. )$\lambda = \frac{2mc}{2h}\lambda'^{2}$
• C. )$\lambda = \frac{2mc}{h}\lambda'^{2}$
• D. )$\lambda^{'} = \frac{mc}{h}\lambda'^{2}$

Answer 1

Answer: (C)

)$\lambda = \frac{2mc}{h}\lambda'^{2}$

Answer 2

: kinetic energy of emitted electron= Energy of incident photon

$i.e.\frac{1}{2}mv^{2} = h\upsilon$

Or $\frac{p^{2}}{2m} = \frac{hc}{\lambda}\left( \because mv = p,\upsilon = \frac{c}{\lambda} \right)$

Or $p = \sqrt{\frac{2mhc}{\lambda}}$

de Broglie wavelength of emitted electrons

}{or\lambda' = \sqrt{\frac{h\lambda}{2mc}}\therefore\lambda = \frac{2mc}{h}\lambda'^{2}}$$