A particle moving with kinetic energy E has de Broglie wavelength $\lambda$. If energy $\Delta \mathrm{E}$;is added to its energy, the wavelength become $\lambda / 2$. Value of $\Delta \mathrm{E}$, is:

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Modern Physics

A particle moving with kinetic energy E has de Broglie wavelength

λ

. If energy

Δ E

;is added to its energy, the wavelength become

λ /2

. Value of

Δ E

, is:

(A)E
(B)3 E✓ Correct
(C)2 E
(D)4 E

Answer: (B)

Solution

For a non-relativistic particle the de Broglie wavelength is tied to kinetic energy by

λ = \frac{h}{2 m E}

, so

λ \propto \frac{1}{E}

. Halving the wavelength therefore means

E

has doubled, i.e. the energy has grown fourfold. Writing

\frac{λ}{λ /2} = 2 = \frac{E + Δ E}{E}

and squaring gives

\frac{E + Δ E}{E} = 4

, so

Δ E = 3 E

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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