JEE Main · 2006 · 30 Apr 2006 · Offline · Modern Physics

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The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in

(A)X-ray region
(B)ultra-violet region✓ Correct
(C)infra-red region
(D)visible region

Answer: (B)

Solution

By Einstein's photoelectric equation the photon energy is the work function plus the maximum kinetic energy of the ejected electron,

E = ϕ + e V_{s} = 6.2 + 5 = 11.2 eV

. The corresponding wavelength is

λ = \frac{1240 eV nm}{11.2 eV} \approx 111 nm

, which is shorter than any visible wavelength and lies in the ultraviolet region.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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