MHT CET · Physics · Thermal Properties of Matter
A black body has maximum wavelength \(\lambda_{\mathrm{m}}\) at temperature \(2200 \mathrm{~K}\). Its
corresponding wavelength at temperature \(3300 \mathrm{~K}\) will be
- A \(\frac{9}{4} \lambda_{\mathrm{m}}\)
- B \(\frac{3}{2} \lambda_{\mathrm{m}}\)
- C \(\frac{4}{9} \lambda_{\mathrm{m}}\)
- D \(\frac{2}{3} \lambda_{\mathrm{m}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{3} \lambda_{\mathrm{m}}\)
Step-by-step Solution
Detailed explanation
\(\lambda_{\mathrm{m}} \mathrm{T}=\) constant
\(\therefore \frac{\lambda_{\mathrm{m}}}{\lambda_{\mathrm{m}}}=\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{2200}{3300}=\frac{2}{3}\)
\(\therefore \lambda_{\mathrm{m}}^{\prime}=\frac{2}{3} \lambda_{\mathrm{m}}\)
\(\therefore \frac{\lambda_{\mathrm{m}}}{\lambda_{\mathrm{m}}}=\frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{2200}{3300}=\frac{2}{3}\)
\(\therefore \lambda_{\mathrm{m}}^{\prime}=\frac{2}{3} \lambda_{\mathrm{m}}\)
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