Question

Threshold wavelength of a metal is ${\lambda }_0$. The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength $\lambda$ is:

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

Answer 1

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

$Explanation:$

From $Einstei{n}^{\prime }sphotoelectricequation$ we get:

$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_{th}}+K.E$

where,

$\lambda =$Wavelength

${\lambda }_{th}=$threshold wavelength

$h=$Planck's constant

$c=$Velocity of light

$\because K.E=\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_{th}}]$

$DeBrogli{e}^{\prime }swavelength$ attached to it,${\lambda }_d=\frac{h}{p}=\frac{h}{mv}$

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

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

$=[\frac{h^2\lambda {\lambda }_{th}}{2hmc({\lambda }_{th}-\lambda )}{]}^{\frac{1}{2}}=[\frac{h\lambda {\lambda }_{th}}{2mc({\lambda }_{th}-\lambda )}{]}^{\frac{1}{2}}$

$\therefore {\lambda }_d=\left[\frac{h\lambda {\lambda }_{th}}{2mc({\lambda }_{th}-\lambda )}\right]$
where ${\lambda }_{th}={\lambda }_o$

$Hencethecorrectanswerisoption\left(D\right).$