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  • What is the ratio of average thermal energy of Helium atom at room temperature and at the surface of the Sun (6000K)?<img alt="\left(K_{B}=1.38 \times 0^{-23}, {\text {Room Temp. }}=27^{\circ}

What is the ratio of average thermal energy of Helium atom at room temperature and at the surface of the Sun (6000K)?\left(K_{B}=1.38 \times 0^{-23}, {\text {Room Temp. }}=27^{\circ} C)\right.

 

Option: 1

5 \times 10^{-4}


Option: 2

7 \times 10^{-4}


Option: 3

9 \times 10^{-4}


Option: 4

10 \times \mathrm{10}^{-4}


Answers (1)

best_answer

At room temperature, 300K.

Average thermal energy  =\frac{3}{2} \times 1.38 \times 10^{-23} \times 300

=6.2 \times 10^{-23} \mathrm{~J}

Average thermal energy at Sun (6000K) =\frac{3}{2} \times 1.38 \times 10^{-23} \times 6000

\begin{aligned} & =1.248 \times 10^{-19} \mathrm{~J} \\ \text { Ratio } & =\frac{6.2 \times 10^{-23}}{1.24 \times 10^{-19}} \\ & =5 \times 10^{-4} \end{aligned}

 

 

Posted by

Nehul

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