Question

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

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

Answer 1

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

$hv-W_0=\frac{1}{2}m{v}_{max}^2\Rightarrow \frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{1}{2}m{v}_{max}^2$
$\Rightarrow hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)=\frac{1}{2}m{v}_{max}^2\Rightarrow v_{max}=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$
When wavelength is $\lambda$ and velocity is v, then
$v=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)}$ …. (i)
When wavelength is $\frac{3\lambda }{4}$ and velocity is v' then
$v^{\prime }=\sqrt{\frac{2hc}{m}\left[\frac{{\lambda }_0-\left(\frac{3\lambda }{4}\right)}{\left(\frac{3\lambda }{4}\right)\times {\lambda }_0}\right]}$ ….(ii)
Divide equation (ii) by (i), we get
$\frac{v^{\prime }}{v}=\sqrt{\frac{[{\lambda }_0-\left(\frac{3\lambda }{4}\right)]}{\frac{3}{4}\lambda {\lambda }_0}\times \frac{\lambda {\lambda }_0}{{\lambda }_0-\lambda }}$
$v^{\prime }=v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{[{\lambda }_0-\left(\frac{3\lambda }{4}\right)]}{{\lambda }_0-\lambda }}i.e.v^{\prime }>v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$