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JEE Main 2022 · 26 Jun · Shift 2, Physics question 47

A metal surface is illuminated by a radiation of wavelength 4500 Å. 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 :

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Answer and worked solution

Official answer: A

Electron speed. v=eBrm=1.76×1011×2×10−3×2×10−3≈7.0×105v=\dfrac{eBr}m=1.76\times10^{11}\times2\times10^{-3}\times2\times10^{-3}\approx7.0\times10^5 m s−1^{-1}, so K=12mv2≈1.4K=\frac12mv^2\approx1.4 eV.

Photon energy. 124004500≈2.76\dfrac{12400}{4500}\approx2.76 eV.

Work function. 2.76−1.40≈1.362.76-1.40\approx1.36 eV.

(A)

Topics:Photoelectric effectMotion in magnetic and electric fields
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