Question

Photons of energy $2.4eV$ and wavelength $\lambda$ fall on a metal plate and release photo electrons with a maximum velocity $v$. By increase in $\lambda$ by $50\%$ the maximum velocity of photoelectrons becomes $3v$. The velocity if the material of the metal plate is

• A. $2.1eV$
• B. $1.7eV$
• C. $2.8eV$
• D. $2.0eV$

Answer 1

Answer: (A)

Let phi ${}_0=$ work-function of metal.

Then, according to Einstein's photoelectric equation,
$K_{\max }=\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-{\phi }_0...(i)$
When wavelength is reduced by $50\%$
i.e, ${\lambda }^{\prime }=\frac{\lambda }{2}$, then maximum velocity of emitted electrons is $3v$.
$\Rightarrow \frac{1}{2}m(3v{)}^2=\frac{hc}{\frac{\lambda }{2}}-{\phi }_0$
$\Rightarrow 9\left(\frac{1}{2}m{v}^2\right)=\frac{2hc}{\lambda }-{\phi }_0\dots (ii)$
From Eqs. (i) and (ii), we have
$9\left(\frac{hc}{\lambda }-{\phi }_0\right)=\frac{2hc}{\lambda }-{\phi }_0$
$7\frac{hc}{\lambda }=8{\phi }_0\Rightarrow {\phi }_0=\frac{7}{8}\frac{hc}{\lambda }$
Here, photon energy is given as $\frac{hc}{\lambda }=2.4eV$
So, work-function is
${\phi }_0=\frac{7}{8}\times 2.4=2.1eV$