Question

In a photoemissive cell with exciting wave length $\lambda$, the fastest electron has a speed $v$. If the exciting wavelength is change to $3\lambda /4$, then 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 $v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

From Einstein's photoelectric equation,
$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W....\left(i\right)$

where $W$ is the work function of metal.
For case 2 :

$\frac{1}{2}m(v^{\prime }{)}^2=\frac{hc}{3\lambda /4}-W....(ii)$
On dividing Eq. (ii) by Eq. (i), we get
${\left(\frac{v^{\prime }}{v}\right)}^2=\frac{\frac{4hc}{3\lambda }-W}{\frac{hc}{\lambda }-W}$
$\because W>0$
$⟹v^{\prime }>v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$.