The wavelength of maximum energy, released during an atomic explosion was 2.93 × 10-10 m. Given that the Wien's constant is 2.93 × 10-3 m - K, the maximum temperature attained must be of the order of :

Question

The wavelength of maximum energy, released during an atomic explosion was 2.93 × 10-10 m. Given that the Wien's constant is 2.93 × 10-3 m - K, the maximum temperature attained must be of the order of : (A) 10-7K (B) 107K (C) 10-3K (D) 5.86 × 107K

Tardigrade Admin · Accepted Answer

Wien's displacement law is given by λm T= constant (say b). Given, b = Wien's constant =2.93 × 10-3 m - K λm =2.93 × 10-10 m Substituting the values, we obtain T =(b/λm) =(2.93 × 10-3/2.93 × 10-10) =107 K

Q. The wavelength of maximum energy, released during an atomic explosion was $2.93 \times 1 0^{- 10} m$ . Given that the Wien's constant is $2.93 \times 1 0^{- 3} m - K$ , the maximum temperature attained must be of the order of :

$1 0^{- 7} K$

$1 0^{7} K$

$1 0^{- 3} K$

$5.86 \times 1 0^{7} K$

Wien's displacement law is given by
$λ_{m} T =$ constant (say b).
Given, $b =$ Wien's constant
$= 2.93 \times 1 0^{- 3} m - K$
$λ_{m} = 2.93 \times 1 0^{- 10} m$
Substituting the values, we obtain
$T = \frac{b}{λ _{m}}$
$= \frac{2.93 \times 1 0 ^{- 3}}{2.93 \times 1 0 ^{- 10}}$
$= 1 0^{7} K$

Q. The wavelength of maximum energy, released during an atomic explosion was 2.93×10−10m. Given that the Wien's constant is 2.93×10−3m−K, the maximum temperature attained must be of the order of :

10−7K

107K

10−3K

5.86×107K