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Physics Question on thermal properties of matter

The wavelength of maximum energy, released during an atomic explosion was 2.93×10−10m2.93 \times 10^{-10} m. Given that the Wien's constant is 2.93×10−3m−K2.93 \times 10^{-3} m - K, the maximum temperature attained must be of the order of :

A

10−7K10^{-7}K

B

107K10^{7}K

C

10−3K10^{-3}K

D

5.86 ×107K5.86\, \times 10^{7}K

Answer

107K10^{7}K

Explanation

Solution

Wien's displacement law is given by
λmT=\lambda_{m} T= constant (say b).
Given, b=b = Wien's constant
=2.93×10−3m−K=2.93 \times 10^{-3} m - K
λm=2.93×10−10m\lambda_{m} =2.93 \times 10^{-10} m
Substituting the values, we obtain
T=bλmT =\frac{b}{\lambda_{m}}
=2.93×10−32.93×10−10=\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}
=107K=10^{7} K