SATHEE: dual-nature-of-radiation-and-matter Question 49

Dual Nature Of Radiation And Matter

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dual-nature-of-radiation-and-matter Question 49

dual-nature-of-radiation-and-matter Question 49

Question: Q. 6. The wavelength $λ$ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon is $(2 λ m c / h)$ times the kinetic energy of electron; where $m, c$ and $h$ have their usual meaning. [Foreign I, II, III 2016]

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Solution:

Ans. Energy of photon $E = h v = \frac{h c}{λ}$

$\Rightarrow \frac{h}{λ} = \frac{E}{c}$

de-Broglie wavelength of electron

$λ = \frac{h}{p}$

Kinetic energy of electron,

$K = \frac{p^{2}}{2 m}$

$\begin{aligned} = \frac{h^{2}}{2 m λ^{2}} = (\frac{h}{2 m λ}) (\frac{h}{λ}) & = (\frac{h}{2 m λ}) (\frac{E}{c}) \Rightarrow E & = (\frac{2 m c λ}{h}) K \end{aligned}$