Question

A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity v m / s, the cut off wavelength being 600 nm The same surface would release electrons with maximum velocity 2\, v m / s if it is irradiated by light of wavelength

• A. 325 nm
• B. 360 nm
• C. 384 nm
• D. 300 nm

Answer 1

Answer: (D)

According to Einstein's photoelectric equation
$\frac{1}{2} m v_{\max }^{2}=h v-h v_{0} $ or
$ \frac{1}{2} m v_{\max }^{2}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}$
where $\lambda$ is the wavelength of incident radiation
and $\lambda_{0}$ is threshold wavelength.
$\therefore \frac{1}{2} m v^{2}=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right) \dots$(i)
$ \frac{1}{2} m(2 v)^{2}=h c\left(\frac{1}{\lambda'}-\frac{1}{\lambda_{0}}\right) \dots$(ii)
Divide (i) by (ii), we get
$\frac{1}{4}=\frac{\frac{1}{\lambda}-\frac{1}{\lambda_{0}}}{\frac{1}{\lambda'}-\frac{1}{\lambda_{0}}} $ or
$ \frac{1}{4}=\frac{\frac{1}{480}-\frac{1}{600}}{\frac{1}{\lambda^{\prime}}-\frac{1}{600}}$
Solving for $\lambda'$, we get $\lambda'=300 \,nm$