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Dual Nature of Radiation question

2023 · 10 Apr · Shift 1 · Q55

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Dual Nature of Radiation
2023 · 10 Apr · Shift 1 · Q55

Dual Nature of Radiation question

2023 · 10 Apr · Shift 1 · Q55

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is

\lambda_1

. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

A
2 $\lambda_1$
B
$\frac{1}{2}\lambda_1$
C
$\frac{1}{\sqrt2}\lambda_1$
D
$\sqrt2~\lambda_1$

View written solutionFree

Correct answer: C

Use de Broglie relation

For a particle,

\lambda = \frac{h}{p}

where $h$ is Planck’s constant and $p$ is momentum.

Relate momentum to temperature for a gas molecule

For a gas molecule at temperature $T$ , the average kinetic energy is proportional to temperature:

\frac{p^2}{2m} \propto T

So,

p^2 \propto T \quad \Rightarrow \quad p \propto \sqrt{T}

Hence wavelength dependence on temperature

Since

\lambda = \frac{h}{p},

and $p \propto \sqrt{T}$ ,

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

Compare wavelengths at $300\,\text{K}$ and $600\,\text{K}$

Given:

At $T_1 = 300\,\text{K}$ , wavelength $= \lambda_1$
At $T_2 = 600\,\text{K}$ , wavelength $= \lambda_2$

Therefore,

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

So,

\lambda_2 = \frac{1}{\sqrt{2}}\lambda_1

Check options

A: $2\lambda_1$ ❌
B: $\frac{1}{2}\lambda_1$ ❌
C: $\frac{1}{\sqrt2}\lambda_1$ ✅
D: $\sqrt2\lambda_1$ ❌

Therefore, the correct option is C.

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