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JEE Main 2019 (10 Apr, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2019 (10 Apr, Shift 1)

Q129JEE Main 2019 · 10 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation

In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be: Given E (in eV) = 1237/ $\lambda$ (in nm)

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Answer: (d)

$K_{max} = \frac{1237}{\lambda} - \frac{1237}{\lambda_0}$ (in eV, $\lambda$ in nm)
$K_{max} = 1237\left(\frac{1}{260} - \frac{1}{380}\right) = 1237 \times \frac{120}{98800}$
$K_{max} \approx 1.5\text{ eV}$ , so the answer is (d) 1.5 eV.

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