Red light, however bright, cannot produce the emission of electrons from a clean zinc surface, but even weak ultraviolet radiation can do so. Why? X-rays of wavelength $\lambda$ fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of the electrons emitted will be $\sqrt{\dfrac{h\lambda}{2mc}}$.

Red is below zinc's threshold frequency; $\lambda_e = \sqrt{h\lambda/2mc}$.

CBSE Class 12 · 2004 · Modern PhysicsShort Answer

Red light, however bright, cannot produce the emission of electrons from a clean zinc surface, but even weak ultraviolet radiation can do so. Why? X-rays of wavelength

λ

fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of the electrons emitted will be

\frac{hλ}{2 m c}

Answer: Red is below zinc's threshold frequency; $λ_{e} = hλ /2 m c$ .

Solution

Photoemission occurs only when the frequency exceeds the threshold frequency of the metal. Red light has a frequency below the threshold of zinc, so it cannot emit electrons however intense it is; ultraviolet light has a frequency above the threshold, so even a weak beam causes emission.
For the X-ray part, neglecting the work function, all the photon energy becomes kinetic energy:

\frac{h c}{λ} = \frac{1}{2} m v^{2} \Rightarrow m v = \frac{2 mh c}{λ} .

The de-Broglie wavelength of the emitted electron is

λ_{e} = \frac{h}{m v} = \frac{h}{2 mh c / λ} = \frac{h ^{2} λ}{2 mh c} = \frac{hλ}{2 m c} .

λ_{e} = \frac{hλ}{2 m c}

CBSE Class 12 Board · 2004 · Subjective

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CBSE Class 12 · 2004 · Modern PhysicsShort Answer

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