Question

Photons of energy $2.4\, eV$ 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 $3\, v$. The velocity if the material of the metal plate is

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

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{h c}{\lambda}-\varphi_{0} \,\, ...(i) $
When wavelength is reduced by $50 \%$
i.e, $\lambda'=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.
$\Rightarrow \frac{1}{2} m(3 v)^{2}=\frac{h c}{\frac{\lambda}{2}}-\varphi_{0} $
$\Rightarrow 9\left(\frac{1}{2} m v^{2}\right)=\frac{2 h c}{\lambda}-\varphi_{0} \ldots(i i)$
From Eqs. (i) and (ii), we have
$9\left(\frac{h c}{\lambda}-\varphi_{0}\right)=\frac{2 h c}{\lambda}-\varphi_{0}$
$7 \frac{h c}{\lambda}=8 \varphi_{0} \Rightarrow \varphi_{0}=\frac{7}{8} \frac{h c}{\lambda}$
Here, photon energy is given as $\frac{h c}{\lambda}=2.4\, eV$
So, work-function is
$\varphi_{0}=\frac{7}{8} \times 2.4=2.1\,eV$