Two positively charged particles $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of $\mathrm{m}_{1}=1$ amu and $\mathrm{m}_{2}=4$ amu] The deBroglie wavelength of $\mathrm{m}_{1}$ will be x times of $\mathrm{m}_{2}$. The value of x is ______. (nearest integer)

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Atomic Structure

Two positively charged particles

m_{1}

and

m_{2}

have been accelerated across the same potential difference of 200 keV as shown below.

[Given mass of

m_{1} = 1

amu and

m_{2} = 4

amu]
The deBroglie wavelength of

m_{1}

will be x times of

m_{2}

. The value of x is ______. (nearest integer)

Answer: 2

Solution

Accelerating charges of the same sign through the same potential difference gives both particles the same kinetic energy, and the de Broglie wavelength is

λ = \frac{h}{2 m E}

, so

λ \propto \frac{1}{m}

at fixed

E

. Then

\frac{λ _{m_{1}}}{λ _{m_{2}}} = \frac{m _{2}}{m _{1}} = \frac{4}{1} = 2

. The wavelength of

m_{1}

is 2 times that of

m_{2}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer

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