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
$\frac{1}{2}m{v}_{\max }^2=hv-h{v}_0$ or
$\frac{1}{2}m{v}_{\max }^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
where $\lambda$ is the wavelength of incident radiation
and ${\lambda }_0$ is threshold wavelength.
$\therefore \frac{1}{2}m{v}^2=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)\dots$(i)
$\frac{1}{2}m(2v{)}^2=hc\left(\frac{1}{{\lambda }^{\prime }}-\frac{1}{{\lambda }_0}\right)\dots$(ii)
Divide (i) by (ii), we get
$\frac{1}{4}=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }_0}}{\frac{1}{{\lambda }^{\prime }}-\frac{1}{{\lambda }_0}}$ or
$\frac{1}{4}=\frac{\frac{1}{480}-\frac{1}{600}}{\frac{1}{{\lambda }^{\prime }}-\frac{1}{600}}$
Solving for ${\lambda }^{\prime }$, we get ${\lambda }^{\prime }=300nm$