JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2 E$ , is $n : 1$ . The value of $n$ is ____. (Assume mass of proton = mass of neutron) :

Your answer

Solution

De Broglie wavelength: $λ = \frac{h}{p} = \frac{h}{2 m K}$ .

Deuteron: mass $m_{d} = 2 m_{p}$ , KE $= E$ .

Alpha: mass $m_{α} = 4 m_{p}$ , KE $= 2 E$ .

$\frac{λ _{d}}{λ _{α}} = \frac{m _{α} \cdot 2 E}{m _{d} \cdot E} = \frac{4 m _{p} \cdot 2 E}{2 m _{p} \cdot E} = 4 = 2$ .

$∴ n = 2$ .

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