SolveIt is under development
SolveItJEE Main
Question 152 / 153 · latest first
Physics · 2018

JEE Main · 15 April 2018, Shift 2 · Q11

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 ]

The value closest to the thermal velocity of a Helium atom at room temperature ($\displaystyle 300$ K) in $\displaystyle \mathrm{ms}^{-1}$ is : $\displaystyle \left[\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right.$; $\displaystyle \mathrm{m}_{\mathrm{He}}=7 \times 10^{-27} \mathrm{~kg}$ ]
ShareWhatsAppTelegram

More from Kinetic Theory of Gases

JEE Main 2018 Physics question, with the answer from NTA’s final answer key. Where our answers come from.