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Question

Physics Question on kinetic theory

Approximately, the temperature corresponding to 1 eV energy, is

A

7.6×102K7.6 \times 10^{2} K

B

7.7×103K7.7 \times 10^{3} K

C

7.1×103K7.1 \times 10^{3} K

D

7.2×103K7.2 \times 10^{3} K

Answer

7.7×103K7.7 \times 10^{3} K

Explanation

Solution

Given: Energy (E)=1eV=1.6×1019J(E)=1 eV =1.6 \times 10^{-19} \,J. Average kinetic energy per molecule (E)=1.6×1019J(E)=1.6 \times 10^{-19} \,J =32kT=\frac{3}{2}\, k T or T=2×(1.6×1019)3kT=\frac{2 \times\left(1.6 \times 10^{-19}\right)}{3 k} =2×(1.6×1010)3×(138×1021)=\frac{2 \times\left(1.6 \times 10^{-10}\right)}{3 \times\left(1 \cdot 38 \times 10^{-21}\right)} =7.7×10K=7.7 \times 10' K (where kk is Boltzman constant =1.38×1023=1.38 \times 10^{-23}.