Question

The de Broglie wavelength of a molecule in a gas at room tem perature (300 K) is λ₁. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

• A. \(\frac{\lambda}{2}\)
• B. \({2\lambda}\)
• C. \({\lambda}\)
• D. \(\frac{5\lambda}{2}\)

Hint:

The de Broglie wavelength is inversely proportional to the square root of the kinetic energy, which is directly proportional to the temperature

Answer 1

The Correct Option is A

Calculation Using Wien’s Displacement Law:

Step 1: Formula

According to Wien’s displacement law:

\( \lambda \propto \frac{1}{T} \)

or equivalently:

\( \lambda T = \text{constant} \)

Thus, for two states:

\( \lambda_1 T_1 = \lambda_2 T_2 \)

Step 2: Given Data

• Initial temperature: \( T_1 = 300 \, \text{K} \)
• Initial wavelength: \( \lambda_1 = \lambda \)
• Temperature increase: \( \Delta T = 300 \, \text{K} \)
• Final temperature: \( T_2 = T_1 + \Delta T = 300 + 300 = 600 \, \text{K} \)

Step 3: Calculate New Wavelength

From the formula \( \lambda_1 T_1 = \lambda_2 T_2 \):

\[ \lambda_2 = \frac{\lambda_1 T_1}{T_2} \]

Substitute the values:

\[ \lambda_2 = \frac{\lambda \cdot 300}{600} \]

Simplify:

\[ \lambda_2 = \frac{\lambda}{2} \]

Conclusion:

The new wavelength corresponding to the maximum intensity is \( \frac{\lambda}{2} \).

Answer: A

Answer 2

\(\lambda T\) = constant

\(\therefore\) \(\frac{\lambda_1}{\lambda_2}=\frac{T_2}{T_1}\)

\(\lambda_2=\lambda(\frac{300}{600})=\frac{\lambda}{2}\)