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प्रश्न
The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon in (2λmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning.
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उत्तर १
Given that λ is the wavelength of the photon.
The de-Broglie wavelength of the electron is \[\lambda_e = \frac{h}{mv}\].
The kinetic energy of the electron is given as
\[E_e = \frac{1}{2}m v^2 = \frac{1}{2}m(\frac{h}{m \lambda_e} )^2 = \frac{h^2}{2m \lambda_e^2} \]
[∵ according to the question λ = λe]
We know that the energy of the photon is
उत्तर २
As, `lambda = h/(mv), v = h/(mlambda)` ...(i)
Energy of photon `E = (hc)/lambda`
& Kinetic energy of electron
`K = 1/2mv^2 = 1/2(mh^2)/(m^2lambda^2)` ...(ii)
Simplifying equation (i) & (ii) we get,
`E/K = (2λmc)/h`
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