The de-Broglie wavelength of a neutron at 27^{} C is ' λ₀ '. What will be its wavelength at 927°C?

Correct Answer is : λ02{λ₀}{2}2λ0

Correct Answer is : λ02{λ₀}{2}2λ0

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 22 of 150

Marks: +1, -0

27^{\circ} C

\lambda_{0}

'. What will be its wavelength at 927°C?

Solution:

Concept:

At a given temperature, the de-Broglie wavelength of a thermal neutron is inversely proportional to the square root of its absolute temperature.

Explanation:

The de-Broglie wavelength is given by:

\lambda = \frac{h}{p}

For a neutron in thermal equilibrium, the average kinetic energy is proportional to the absolute temperature:

\frac{1}{2}mv^2 \propto T

So, the velocity and momentum vary as:

v \propto \sqrt{T}, \quad p = mv \propto \sqrt{T}

Hence, the wavelength becomes:

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

Now convert temperatures to Kelvin:

T_1 = 27 + 273 = 300 \, \text{K}

T_2 = 927 + 273 = 1200 \, \text{K}

Taking the ratio of wavelengths:

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Therefore:

\lambda_2 = \frac{\lambda_0}{2}

Answer:

Option C:

\frac{\lambda_0}{2}

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