Question

The surface of certain metal is first illuminated with light of wavelength $\lambda _{1}=350 \, nm$ and then, by a light of wavelength $\lambda _{2}=540 \, nm.$ It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of $2$ . The work function of the metal (in $eV$ ) is close to
(Energy of photon $=\frac{1240}{\lambda \left(i n \, n m\right)} \, eV$ )

• A. $2.5$
• B. $1.8$
• C. $5.6$
• D. $1.4$

Answer 1

Answer: (B)

From Einstein's photoelectric equation, we have,
$K . E_{\max }=\frac{h c}{\lambda}-\phi$
$\Rightarrow \frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi \ldots$ (1), $v$
represents maximum velocity of photoelectrons.
Given,
$\lambda_1=350 nm , \lambda_2=540 nm , \frac{V_1}{V_2}=2 \Rightarrow V _1=2 V$ and $V _2=V$
From equation (1), we have
$\frac{h c}{\lambda_1}=\phi+\frac{1}{2} m(2 v)^2 \ldots(2)$
$\frac{h c}{\lambda_2}=\phi+\frac{1}{2} m v^2 \ldots(3)$
Multiplying equation (2) by 4 and subtracting from equation (1), we have
$ \Rightarrow \frac{h c}{\lambda_1}-4 \frac{h c}{\lambda_2}=\phi-4 \phi $
$ \begin{array}{l} \Rightarrow h c\left(\frac{4}{540}-\frac{1}{340}\right)=3 \phi \\ \Rightarrow \phi=1.8 eV \end{array} $