Question

The surface of certain metal is first illuminated with light of wavelength ${\lambda }_1=350nm$ and then, by a light of wavelength ${\lambda }_2=540nm.$ 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 $=\frac{1240}{\lambda \left(innm\right)}eV$ )

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

Answer 1

Answer: (B)

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