MCQ
The average translational kinetic energy of ${N}_{2}$ gas molecules at $\ldots \ldots \ldots . .{ }^{\circ} {C}$ becomes equal to the ${K} . {E}$. of an electron accelerated from rest through a potential difference of $0.1$ $volt.$

$\left(\right.$ Given $\left.{k}_{{B}}=1.38 \times 10^{-23} \, {J} / {K}\right)$

(Fill the nearest integer).

  • $500$
  • B
    $50$
  • C
    $5$
  • D
    $0.5$

Answer

Correct option: A.
$500$
a
Given

Translation K.E. of ${N}_{2}={KE}$. of electron

$\frac{3}{2} {kT}={eV}$

$\frac{3}{2} \times 1.38 \times 10^{-23} {T}=1.6 \times 10^{-19} \times 0.1$

$\Rightarrow {T}=773 {k}$

${T}=773-273=500^{\circ} {C}$

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