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JEE Main 2019 (9 Jan, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2019 (9 Jan, Shift 1)

Q136JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation

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)

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Answer: (a)

$E_1 = \frac{1240}{350} = 3.54$ eV, $E_2 = \frac{1240}{540} = 2.30$ eV.
$v_1 = 2v_2 \Rightarrow K_1 = 4K_2$ : $3.54 - \phi = 4(2.30 - \phi)$
$3\phi = 9.18 - 3.54 = 5.64 \Rightarrow \phi \approx 1.88$ eV
So the answer is (a) 1.8.

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