Question

If ${\lambda }_0$ is the threshold wavelength for photoelectric emission $\lambda$ wavelength of light falling on the surface of metal, and m mass of electron. Then de Broglie wavelength of emitted electron is :

• A. ${\left[\frac{{h(\lambda \lambda }_0)}{2mc({\lambda }_0-\lambda )}\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{h{(\lambda }_0-\lambda )}{2mc\left(\lambda {\lambda }_0\right)}\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{h\left(\lambda {-\lambda }_0\right)}{2mc\lambda {\lambda }_0}\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{h\lambda {\lambda }_0}{2mc}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is A ${\left[\frac{{h(\lambda \lambda }_0)}{2mc({\lambda }_0-\lambda )}\right]}^{\frac{1}{2}}$

Einstein's photoelectric equation:

$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_o}+1/2m{\nu }^2$

$K_1{E}_1=\frac{1}{2}m{\nu }^2=hc(\frac{1}{\lambda }-\frac{1}{\lambda })$

De Broglie's wavelength attached to it, ${\lambda }_{\alpha }=\frac{h}{p}=\frac{h}{mv}$

$\therefore {\lambda }_d=\frac{h}{\sqrt{2K.E\times m}}=\frac{h}{\sqrt{2\times hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_o})\times m}}$

$=\frac{h}{\sqrt{2\times hmc\times \left(\frac{{\lambda }_o-\lambda }{\lambda {\lambda }_o}\right)}}$

$={\left[\frac{h^2\lambda {\lambda }_o}{2hmc({\lambda }_o-\lambda )}\right]}^{1/2}={\left[\frac{h\lambda {\lambda }_o}{2mc({\lambda }_o-\lambda )}\right]}^{1/2}$