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
$\text{hν}={\text{hν}}_0+\frac{1}{2}{\text{mv}}_{\text{max}}^2$
$\text{or}\frac{1}{2}{\text{mv}}_{\text{max}}^2=h\nu -h\nu {}_0$
According to the given problem
$\frac{1}{2}m{\left(8\times 10^6\right)}^2=h\left(5\nu {}_0-\nu {}_0\right)\dots \left(\text{i}\right)$
$\frac{1}{2}m{v}_{\text{max}}^2=h\left(2\nu {}_0-\nu {}_0\right)\dots \left(\text{i}\text{i}\right)$
Dividing equation $(i)$ by $(ii)$
$\frac{{\left(8\times 10^6\right)}^2}{v_{\text{max}}^2}=\frac{4{\nu }_0}{\nu {}_0}$
$v_{\text{max}}^2=\frac{{\left(8\times 10^6\right)}^2}{4}$
$v_{\text{max}}=\frac{8\times 10^6}{2}=4\times 10^{6}m{s}^{-1}$