The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field 3 × 10⁻⁴ T. If the radius of the larger circular path followed by these electrons is 10.0 mm, the work function of the metal is close to: (a) 1.8 eV (b) 1.1 eV (c) 0.8 eV (d) 1.6 eV

Answer: (b). Photon energy ( 3 → 2): E = 13.6(1/4 - 1/9) = 1.89 eV. Fastest electrons have the largest radius: p = eBr = 1.6 × 10⁻¹⁹ × 3 × 10⁻⁴ × 0.01 = 4.8 × 10⁻²⁵ kg m/s. Kmax = p²/2m = (4.8 × 10⁻²⁵)²2 × 9.1 × 10⁻³¹ = 1.27 × 10⁻¹⁹ J ≈ 0.79 eV. φ = E - Kmax = 1.89 - 0.79 ≈ 1.1 eV. So the answer is (b) 1.1 eV.

The radiation corresponding to 3 → 2 transition of hydrogen atom… | JEE Main 2014

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Transitions, Photon Emission & Excitation

The radiation corresponding to $3 \rightarrow 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $3 \times 10^{-4}\text{ T}.$ If the radius of the larger circular path followed by these electrons is $10.0\text{ mm}$ , the work function of the metal is close to:

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Answer: (b)

Photon energy ( $3 \to 2$ ): $E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89\text{ eV}$ .
Fastest electrons have the largest radius: $p = eBr = 1.6 \times 10^{-19} \times 3 \times 10^{-4} \times 0.01 = 4.8 \times 10^{-25}$ kg m/s.
$K_{max} = \frac{p^2}{2m} = \frac{(4.8 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} = 1.27 \times 10^{-19}\text{ J} \approx 0.79\text{ eV}$ .
$\phi = E - K_{max} = 1.89 - 0.79 \approx 1.1\text{ eV}$ .
So the answer is (b) $1.1$ eV.

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