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,
$\frac{1}{2} m v^{2}=\frac{h c}{\lambda}-W \ldots$ (i)
and $\frac{1}{2} m v^{' 2}=\frac{h c}{3 \lambda / 4}-W \ldots$ (ii)
On dividing Eq. (ii) by Eq. (i). we get
$\left(\frac{v'}{v}\right)^{2}=\frac{\frac{4 h c}{3 \lambda}-W}{\frac{h c}{\lambda}-W}$
$\because W > 0$
$\therefore v' > v\left(\frac{4}{3}\right)^{\frac{1}{2}}$