Question

In a photoemissive cell with exciting wave length $\lambda$, the fastest electron has a speed $v$. If the exciting wavelength is changed to $\frac{3\lambda }{4}$ 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

Answer: (D)

From Einstein's photoelectric equation,
$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W\dots$ (i)
and $\frac{1}{2}m{v}^{{}^{\prime }2}=\frac{hc}{3\lambda /4}-W\dots$ (ii)
On dividing Eq. (ii) by Eq. (i). we get
${\left(\frac{v^{\prime }}{v}\right)}^2=\frac{\frac{4hc}{3\lambda }-W}{\frac{hc}{\lambda }-W}$
$\because W>0$
$\therefore v^{\prime }>v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$