Question

In a photoemissive cell with exciting wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda }{4}$, the speed of fastest emitted electron will be :

• A. Greater than $v{\left(\frac{4}{3}\right)}^{1/2}$
• B. $v{\left(\frac{3}{4}\right)}^{1/2}$
• C. $v{\left(\frac{4}{3}\right)}^{1/2}$
• D. Less than $v{\left(\frac{4}{3}\right)}^{1/2}$

Answer 1

The correct option is A Greater than $v{\left(\frac{4}{3}\right)}^{1/2}$

According to photoelectric equation,
$hv-{\omega }_0=\frac{1}{2}m{v}_{max}^2$
$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{1}{2}m{v}_{max}^2$
$hc\left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)=\frac{1}{2}m{v}_{max}^2$
$v_{max}=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)}$
When wavelength is $\lambda$ and velocity is $v$, then
$v=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)}$ .....(i)
When wavelength is $\frac{3\lambda }{4}$ and velocity is $v^{\prime }$, then
$v^{\prime }=\sqrt{\frac{2hc}{m}\left[\frac{{\lambda }_0-\left(3\lambda /4\right)}{\left(3\lambda /4\right)\times {\lambda }_0}\right]}$ .....(ii)
Divide equation (ii) by equation (i), we get
$\frac{v^{\prime }}{v}=\sqrt{\frac{{\lambda }_0-3\lambda /4}{\frac{3\lambda {\lambda }_0}{4}}\times \frac{\lambda {\lambda }_0}{{\lambda }_0-\lambda }}$
$v^{\prime }=v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{[{\lambda }_0-\left(3\lambda /4\right)]}{\lambda {\lambda }_0}}\Rightarrow v^{\prime }>v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$