Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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

  • Option 1)

    1.3 ×10^{^{4}}

  • Option 2)

    1.3 ×10^{^{3}}

  • Option 3)

    1.3 ×10^{5}

  • Option 4)

    1.3 ×10^{^{2}}

 

Answers (2)

best_answer

As we learnt

Root mean square velocity -

V_{rms}= \sqrt{\frac{3RT}{M}}

= \sqrt{\frac{3P}{\rho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

\rho = density

 

 v = \sqrt{\frac{3kT}{m}}=\sqrt{\frac{3*0.2*10^{-23}*300}{4*10^{-27}}} 

v = \sqrt{1.8*10^{6}}=1.3*10^{3}m/s

 


Option 1)

1.3 ×10^{^{4}}

Option 2)

1.3 ×10^{^{3}}

Option 3)

1.3 ×10^{5}

Option 4)

1.3 ×10^{^{2}}

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: Wider gap in NIT cut-offs for home and other-state quotas; outsiders need better rank
anam-khan

JEE Mains 2026: Use of calculator 'not allowed’ says NTA; cites ‘typographical error' in information bulletin
anam-khan

JEE Main 2026: NIT Warangal attracts best talent with high BTech cut-offs in top, emerging branches

Latest Question