The de-Broglie wavelength of a particle having kinetic energy E is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

(C) $\frac{16}{9} \mathrm{E}$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Modern Physics

The de-Broglie wavelength of a particle having kinetic energy E is

λ

. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to

75%

of the initial value?

(A) $\frac{1}{9} E$
(B)E
(C) $\frac{16}{9} E$
(D) $\frac{7}{9} E$ ✓ Correct

Answer: (D)

Solution

For a particle of kinetic energy

E

λ = \frac{h}{p} = \frac{h}{2 m E}

, so

E \propto \frac{1}{λ ^{2}}

.
The new wavelength is

λ^{'} = \frac{3}{4} λ

, hence

E^{'} = E (\frac{λ}{λ ^{'}})^{2} = (\frac{4}{3})^{2} E = \frac{16}{9} E

The energy that must be added is

E^{'} - E = \frac{16}{9} E - E = \frac{7}{9} E

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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