JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

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)

Your answer

Solution

For a charged particle accelerated through a fixed potential, the kinetic energy is the same: $K E = q V$ (same $q$ and $V$ here). $λ_{d} = \frac{h}{2 m \cdot K E}$ , so $λ_{d} \propto \frac{1}{m}$ .

$\frac{( λ _{d} ) _{m_{1}}}{( λ _{d} ) _{m_{2}}} = \frac{m _{2}}{m _{1}} = \frac{4}{1} = 2$ .

$∴ x = 2$ .

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