Question

When light of wavelength ${}^{\prime }{\lambda }^{\prime }$ falls on the surface of a metal with threshold wavelength ${}^{\prime }{\lambda }_0^{\prime }$, an electron of mass 'm' is emitted. The de Broglie wavelength of emitted electron is:

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

Answer 1

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

Given:
$\text{Threshold wavelength of the metal surface}={\lambda }_0\text{Wavelength of the light}=\lambda \text{Mass of the electron}=m$

Kinetic energy (KE) of the emitted electrons is given by:
$KE=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})=hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)$

de Broglie wavelength in terms of kinetic energy= $\frac{h}{\sqrt{2KEm}}$
Substituting the above value of KE, we get:
$\frac{h}{\sqrt{2\times hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)\times m}}={\left[\frac{h\left(\lambda {\lambda }_0\right)}{2mc({\lambda }_0-\lambda )}\right]}^{\frac{1}{2}}$