Question

For certain metal, incident frequency $\nu$ is five times threshold frequency $\nu_{0}$ and the maximum velocity of the photoelectrons is $ \, 8 \, \times \, 10^{6} \, m \, s^{- 1}$ . If incident photon frequency is $2\nu_{0}$ , the maximum velocity of photoelectrons will be

• A. $4\times 10^{6} \, m \, s^{- 1}$
• B. $6\times 10^{6} \, m \, s^{- 1}$
• C. $8\times 10^{6} \, m \, s^{- 1}$
• D. $1\times 10^{6} \, m \, s^{- 1}$

Answer 1

Answer: (A)

According to Einstein's photoelectric equation
$\textit{hν}=\textit{hν}_{0}+\frac{1}{2}\textit{mv}_{\text{max}}^{2}$
$\text{or}\frac{1}{2}\textit{mv}_{\text{max}}^{2}=h \nu⁡-h⁡\nu⁡_{0}$
According to the given problem
$\frac{1}{2}m \left(8 \times 1 0^{6}\right)^{2}=h⁡\left(5 \nu ⁡_{0} - \nu ⁡_{0}\right)\ldots \left(\text{i}\right)$
$\frac{1}{2}m v_{\text{max}}^{2}=h⁡\left(2 \nu ⁡_{0} - \nu ⁡_{0}\right)\ldots \left(\text{i} \text{i}\right)$
Dividing equation $\left(\right.i\left.\right)$ by $\left(\right.ii\left.\right)$
$\frac{\left(8 \times 1 0^{6}\right)^{2}}{v_{\text{max}}^{2}}=\frac{4 \nu _{0}}{\nu ⁡_{0}}$
$v_{\text{max}}^{2}=\frac{\left(8 \times 1 0^{6}\right)^{2}}{4}$
$v_{\textit{max}}=\frac{8 \times 10^{6}}{2}=4\times 10^{6}ms^{- 1}$