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Dual Nature of Radiation question

2025 · 29 Jan · Shift 1 · Q51

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Dual Nature of Radiation
2025 · 29 Jan · Shift 1 · Q51

Dual Nature of Radiation question

2025 · 29 Jan · Shift 1 · Q51

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

If

\lambda

and

K

are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :

A
B
C
D

View written solutionFree

Correct answer: D

For a particle of constant mass $m$ , the de Broglie wavelength is

\lambda = \frac{h}{p}

where $p$ is momentum.

For a non-relativistic particle, kinetic energy is

K = \frac{p^2}{2m}

So,

p = \sqrt{2mK}

Substitute this into the de Broglie relation:

\lambda = \frac{h}{\sqrt{2mK}}

Thus,

\lambda \propto \frac{1}{\sqrt{K}}

Rearranging,

\lambda^2 = \frac{h^2}{2mK}

which gives

K \propto \frac{1}{\lambda^2}

Therefore, the graph between $\lambda$ and $K$ is an inverse-square type curve:

as $K$ increases, $\lambda$ decreases,
the curve is not linear,
it falls rapidly at first and then more slowly.

Hence, the correct graphical representation must be the one showing

\lambda \propto \frac{1}{\sqrt{K}}

or equivalently

K \propto \frac{1}{\lambda^2}

which corresponds to Option D.

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