Search any question & find its solution
Question:
Answered & Verified by Expert
If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through $10 \mathrm{~V}$, then the temperature of the gas molecule is
$$
\left(\text { Boltzmann constant }=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)
$$
Options:
$$
\left(\text { Boltzmann constant }=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)
$$
Solution:
1370 Upvotes
Verified Answer
The correct answer is:
$77.3 \times 10^3 \mathrm{~K}$
According to question, the translation $\mathrm{KE}$ of a molecule of gas $=\frac{3}{2} k t$
According to the question,
$$
\begin{aligned}
& \frac{3}{2} k t=e V \Rightarrow \frac{3}{2} k t=10 e \\
& \begin{aligned}
\Rightarrow \quad T & =\frac{20 e}{3 K_B}=\frac{20 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} \\
& =77.3 \times 10^3 \mathrm{~K}
\end{aligned}
\end{aligned}
$$
According to the question,
$$
\begin{aligned}
& \frac{3}{2} k t=e V \Rightarrow \frac{3}{2} k t=10 e \\
& \begin{aligned}
\Rightarrow \quad T & =\frac{20 e}{3 K_B}=\frac{20 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} \\
& =77.3 \times 10^3 \mathrm{~K}
\end{aligned}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.