Solveeit Logo
Login

Question

Physics Question on communication systems

The value closest to the thermal velocity of a Helium atom at room temperature (300 K)(300\, K) in ms−1ms^{-1} is :[kB=1.4×10−23 J/K;mHe=7×10−27 kg] [k_B = 1.4 \times 10^{-23} \, J/K; m_{He} = 7 \times 10^{-27} \, kg ]

A

1.3×1041.3 \times 10^4

B

1.3×1031.3 \times 10^3

C

1.3×1051.3 \times 10^5

D

1.3×1021.3 \times 10^2

Answer

1.3×1031.3 \times 10^3

Explanation

Solution

We know that vrms=3kTmv_{ rms }=\sqrt{\frac{3 k T}{m}}. Given, k=1.4×10−23 J/K;T=300 K;m=7×10−27 kgk=1.4 \times 10^{-23} \,J / K ; T=300 \,K ; m=7 \times 10^{-27}\, kg. Therefore, vrms=3×1.4×10−23×3007×10−27=3×300×14×10−237×10×10−27=32×102×2×10410v_{ rms } =\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}}=\sqrt{\frac{3 \times 300 \times 14 \times 10^{-23}}{7 \times 10 \times 10^{-27}}}=\sqrt{\frac{3^{2} \times 10^{2} \times 2 \times 10^{4}}{10}} =3×10×102×210=3 \times 10 \times 10^{2} \times \sqrt{\frac{2}{10}} =3×10×102×2=3×2×5×102×2=3×2×5×102=3 \times \sqrt{10} \times 10^{2} \times \sqrt{2}=3 \times \sqrt{2} \times \sqrt{5} \times 10^{2} \times \sqrt{2}=3 \times 2 \times \sqrt{5} \times 10^{2} vrms=6×102×5=13.41×102v_{ rms } =6 \times 10^{2} \times \sqrt{5}=13.41 \times 10^{2} or vrms=1.34×103 v_{ rms }=1.34 \times 10^{3}