The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 × 10 - 4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is: (Mass of electron = 9.1 × 10 - 31 kg) (a) 1.36 eV (b) 1.88 eV (c) 0.82 eV (d) 0.16 eV

Answer: (c). 3 → 2: E = 13.6(1/4 - 1/9) = 1.89 eV. K = (eB²r²)/(2m) eV = 1.6 × 10⁻¹⁹ × (5 × 10⁻⁴)² (7 × 10⁻³)²2 × 9.1 × 10⁻³¹ ≈ 1.08 eV φ = 1.89 - 1.08 ≈ 0.81 eV So the answer is (c) 0.82 eV.

The radiation corresponding to 3 → 2 transition of a hydrogen atom… | JEE Main 2021 (20 Jul, Shift 1)

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Photoelectric Effect: Einstein's Equation

The radiation corresponding to 3 $\rightarrow$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 $\times$ 10 $-$ 4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is: (Mass of electron = 9.1 $\times$ 10 $-$ 31 kg)

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Answer: (c)

$3 \to 2$ : $E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89$ eV.
$K = \frac{eB^2r^2}{2m}$ eV $= \frac{1.6 \times 10^{-19} \times (5 \times 10^{-4})^2 (7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.08$ eV
$\phi = 1.89 - 1.08 \approx 0.81$ eV
So the answer is (c) 0.82 eV.

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