JEE PYQ: Atomic Physics Question 58
Question 58 - 2019 (09 Jan Shift 1)
Surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350$ nm and then, by light of wavelength $\lambda_2 = 540$ nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of (2). The work function of the metal (in eV) is close to:
(Energy of photon $= \frac{1240}{\lambda (\text{in nm})}$ eV)
(1) 1.8
(2) 2.5
(3) 5.6
(4) 1.4
Show Answer
Answer: (1)
Solution
From Einstein’s photoelectric equation, $\frac{hc}{\lambda_1} = \phi + \frac{1}{2}m(2v)^2$ …(i) and $\frac{hc}{\lambda_2} = \phi + \frac{1}{2}mv^2$ …(ii). $\frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} = 4$. $\frac{4hc}{\lambda_2} - \frac{hc}{\lambda_1} = 3\phi$. $\phi = \frac{1}{3} \times 1240\left(\frac{4}{540} - \frac{1}{350}\right) = \frac{1}{3} \times 1240 \times \frac{4 \times 350 - 540}{350 \times 540} = 1.8$ eV.
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