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. ${\left[\frac{h\lambda {\lambda }_0}{2mc({\lambda }_0-\lambda )}\right]}^{0.5}$
• B. ${\left[\frac{h({\lambda }_0-\lambda )}{2mc\lambda {\lambda }_0}\right]}^{0.5}$
• C. ${\left[\frac{h(\lambda -{\lambda }_0)}{2mc\lambda {\lambda }_0}\right]}^{0.5}$
• D. ${\left[\frac{h\lambda {\lambda }_0}{2mc}\right]}^{0.5}$

Answer 1

The correct option is A ${\left[\frac{h\lambda {\lambda }_0}{2mc({\lambda }_0-\lambda )}\right]}^{0.5}$

$$\begin{array}{c}Accorrdingtoquestion.......................\\ Here,\\ E=E^0+\frac{1}{2}m{v}^2\\ \Rightarrow \frac{hc}{\lambda }=\frac{hc}{{\lambda }^0}+\frac{1}{2}m{v}^2\\ \Rightarrow \frac{1}{2}m{v}^2=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^0}\right)\\ \Rightarrow m{v}^2=2hc\left(\frac{{\lambda }^0-\lambda }{\lambda .{\lambda }^0}\right)\\ \therefore m^2{v}^2=2hmc\left(\frac{{\lambda }^0-\lambda }{\lambda .{\lambda }^0}\right)---------\left(1\right)\\ Now,\\ Debrogliewavelength,\lambda =\frac{h}{mv}\\ eequationfrom\left(1\right)................\\ \lambda =\frac{h}{\left(2hmc\right){\left(\frac{{\lambda }^0-\lambda }{\lambda .{\lambda }^0}\right)}^{\frac{1}{2}}}\\ \therefore \lambda ={\left(\frac{h\lambda {\lambda }^0}{2mc({\lambda }^0-\lambda )}\right)}^{0.5}\\ SothecorrectoptionisA.\end{array}$$