The de Broglie wavelength for an electron accelerated through the potential difference $V_{1}$ volt is $\lambda_{1}$. When the potential difference is changed to $\mathrm{V}_{2}$ volt, the associated de Broglie wavelength is increased by $50\%$. If $\left(\mathrm{V}_{1} / \mathrm{V}_{2}\right)=(9 / \alpha)$, then the value of $\alpha$ is ______.

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Modern Physics

The de Broglie wavelength for an electron accelerated through the potential difference

V_{1}

volt is

λ_{1}

. When the potential difference is changed to

V_{2}

volt, the associated de Broglie wavelength is increased by

50%

. If

(V_{1} / V_{2}) = (9/ α)

, then the value of

α

is ______.

Answer: 4

Solution

The de Broglie wavelength of an accelerated electron is

λ = \frac{h}{2 m e V}

, so

λ \propto \frac{1}{V}

. A

50%

increase means

λ_{2} = 1.5 λ_{1}

, hence

\frac{λ _{2}}{λ _{1}} = \frac{V _{1}}{V _{2}} = \frac{3}{2}

. Squaring,

\frac{V _{1}}{V _{2}} = \frac{9}{4}

. Comparing with

\frac{9}{α}

gives

α = 4

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer

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