A metal surface is illuminated by a radiation of wavelength 4500. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 ° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately: (a) 1.36 eV (b) 1.69 eV (c) 2.78 eV (d) 2.23 eV

Answer: (a). Photon energy = (12400)/(4500) = 2.76 eV. r = (√(2mK))/(eB) ⇒ K = (e²B²r²)/(2m); in eV, K = (eB²r²)/(2m) = 1.6 × 10⁻¹⁹ × (2 × 10⁻³)² (2 × 10⁻³)²2 × 9.1 × 10⁻³¹ ≈ 1.41 eV φ = 2.76 - 1.41 ≈ 1.35 eV So the answer is (a) 1.36 eV.

A metal surface is illuminated by a radiation of wavelength 4500. The… | JEE Main 2022 (26 Jun, Shift 2)

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Photoelectric Effect: Einstein's Equation

A metal surface is illuminated by a radiation of wavelength 4500 $\text{\AA}$ . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 $^\circ$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately:

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Answer: (a)

Photon energy $= \frac{12400}{4500} = 2.76$ eV.
$r = \frac{\sqrt{2mK}}{eB} \Rightarrow K = \frac{e^2B^2r^2}{2m}$ ; in eV, $K = \frac{eB^2r^2}{2m} = \frac{1.6 \times 10^{-19} \times (2 \times 10^{-3})^2 (2 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 1.41$ eV
$\phi = 2.76 - 1.41 \approx 1.35$ eV
So the answer is (a) 1.36 eV.

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