Question

Threshold wavelength for a metal is ${\lambda }_0$, light of wavelength incident on metal is $\frac{{\lambda }_0}{2}$. Find deBroglie wavelength of electron with maximum K.E. ?

• A. $\frac{2{\lambda }_0}{3}$
• B. $\sqrt{\frac{2h{\lambda }_0}{3mc}}$
• C. $\frac{8{\lambda }_0}{9}$
• D. $\sqrt{\frac{h{\lambda }_0}{2mc}}$

Answer 1

The correct option is D $\sqrt{\frac{h{\lambda }_0}{2mc}}$

$K.E{.}_{\text{max}}=\frac{hc}{\left(\frac{{\lambda }_0}{2}\right)}-\varphi =\frac{2hc}{{\lambda }_0}-\frac{hc}{{\lambda }_0}=\frac{hc}{{\lambda }_0}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mKE}}$

$=\frac{h}{\sqrt{2m.\frac{hc}{{\lambda }_0}}}=\frac{\sqrt{h}}{\sqrt{\frac{2mc}{{\lambda }_0}}}=\sqrt{\frac{h{\lambda }_0}{2mc}}$

$\lambda =\sqrt{\frac{h{\lambda }_0}{2mc}}$