An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $\mathrm{V}_{2}$, its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)$ is equal to :

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Modern Physics

An electron accelerated through a potential difference

V_{1}

has a de-Broglie wavelength of

λ

. When the potential is changed to

V_{2}

, its de-Broglie wavelength increases by

50%

. The value of

(\frac{V _{1}}{V _{2}})

is equal to :

(A)3
(B) $\frac{9}{4}$ ✓ Correct
(C) $\frac{3}{2}$
(D)4

Answer: (B)

Solution

An electron accelerated through a potential difference

V

gains kinetic energy

e V = \frac{p ^{2}}{2 m}

, and the de Broglie relation

p = \frac{h}{λ}

turns this into

e V = \frac{h ^{2}}{2 m λ ^{2}}

, so

λ \propto \frac{1}{V}

and hence

V \propto \frac{1}{λ ^{2}}

. A

50%

rise makes the wavelength

λ_{2} = \frac{3}{2} λ_{1}

, therefore

\frac{V _{1}}{V _{2}} = (\frac{λ _{2}}{λ _{1}})^{2} = (\frac{3}{2})^{2} = \frac{9}{4}

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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