A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly

Dual Nature of Radiation PYQ — JEE Advanced 2014 · Shift 2 · Q47 with Solution

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Correct answer: A

For a given wavelength $\lambda$ , $K_{\max}=\frac{1}{2}mu^2=\frac{hc}{\lambda}-\phi$ where $\phi$ is the work function.

For the two wavelengths: $\frac{1}{2}mu_1^2=\frac{1240}{248}-\phi$ $\frac{1}{2}mu_2^2=\frac{1240}{310}-\phi$

For $\lambda_1=248\,\text{nm}$ : $E_1=\frac{1240}{248}=5\,\text{eV}$

For $\lambda_2=310\,\text{nm}$ : $E_2=\frac{1240}{310}=4\,\text{eV}$

So, $\frac{1}{2}mu_1^2=5-\phi$ $\frac{1}{2}mu_2^2=4-\phi$

Given $u_1:u_2=2:1$ Therefore, $u_1^2:u_2^2=4:1$

Since kinetic energy is proportional to $u^2$ , $\frac{5-\phi}{4-\phi}=4$

$5-\phi=4(4-\phi)$ $5-\phi=16-4\phi$ $3\phi=11$ $\phi=\frac{11}{3}\approx 3.67\,\text{eV}$

So the work function is nearly $\boxed{3.7\,\text{eV}}$

Hence, the correct option is A.