Question

When a certain metal was irradiated with light of frequency $4.0\times {10}^{16}{s}^{-1}$, the photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradicated with light of frequency $2.0\times {10}^{16}{s}^{-1}$. Calculate the critical frequency (${\nu }_0$) of the metal.

• A. $2\times {10}^{16}{s}^{-1}$
• B. $4\times {10}^{16}{s}^{-1}$
• C. $8\times {10}^{16}{s}^{-1}$
• D. $1\times {10}^{16}{s}^{-1}$

Answer 1

The correct option is D

$1\times {10}^{16}{s}^{-1}$

We know that, the incident photons should have a minimum frequency called threshold frequency and a certain amount of energy required called work function. Therefore,

$KE=h\nu -h{\nu }_0=h(\nu -{\nu }_0)$

$K{E}_1=h({\nu }_1-{\nu }_0)$ ...(i)

$K{E}_2=h({\nu }_2-{\nu }_0)$ ...(ii)

Dividing equation (ii) by (i), we get

$\frac{K{E}_2}{K{E}_1}=\frac{h({\nu }_2-{\nu }_0)}{h({\nu }_1-{\nu }_0)}=\frac{({\nu }_2-{\nu }_0)}{({\nu }_1-{\nu }_0)}$

But given that

$\frac{K{E}_2}{K{E}_1}=3$

$\therefore 3=\frac{({\nu }_2-{\nu }_0)}{({\nu }_1-{\nu }_0)}\Rightarrow 3({\nu }_1-{\nu }_0)={\nu }_2-{\nu }_0$

$\Rightarrow 3{\nu }_1-{\nu }_2=3{\nu }_0-{\nu }_0=2{\nu }_0$

$\Rightarrow 3\times 2.0\times {10}^{16}-4\times {10}^{16}=2{\nu }_0$

$\Rightarrow {\nu }_0=\frac{2\times {10}^{16}}{2}=1\times {10}^{16}{s}^{-1}$