Verified Solution • 2025–26 Session moderate

QJEEM18Ph7a00: JEE Main 2018 Physics - Class 11 JEEM18Ph7a00 Solution: The value closest to the thermal velocity of a Helium atom at…

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

Multiple Choice Options

  • A 1.3 ×\times 104
  • B 1.3 ×\times 103
  • C 1.3 ×\times 105
  • D 1.3 ×\times 102

Step-by-Step Verified Solution

We know that vrms=3kBTm{v_{rms}} = \sqrt {{{3{k_B}T} \over m}} . Given, kB=1.4×10−23{k_B} = 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{v_{rms}} = \sqrt {{{3 \times 1.4 \times {{10}^{ - 23}} \times 300} \over {7 \times {{10}^{ - 27}}}}}

=3×300×14×10−237×10×10−27= \sqrt {{{3 \times 300 \times 14 \times {{10}^{ - 23}}} \over {7 \times 10 \times {{10}^{ - 27}}}}}

=32×102×2×10410=3×10×102×210= \sqrt {{{{3^2} \times {{10}^2} \times 2 \times {{10}^4}} \over {10}}} = 3 \times 10 \times {10^2} \times \sqrt {{2 \over {10}}}

=3×10×102×2=3×2×5×102×2= 3 \times \sqrt {10} \times {10^2} \times \sqrt 2 = 3 \times \sqrt 2 \times \sqrt 5 \times {10^2} \times \sqrt 2

=3×2×5×102= 3 \times 2 \times \sqrt 5 \times {10^2}

vrms=6×102×5=13.41×102{v_{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}

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