At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere?
(Given: Mass of oxygen molecule $(\mathrm{m})=2.76 \times 10^{-26} \mathrm{~kg}$ Boltzmann’s constant $\left.k_{B}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$
(1) $5.016 \times 10^{4} \mathrm{~K}$
(2) $8.360 \times 10^{4} \mathrm{~K}$
(3) $2.508 \times 10^{4} \mathrm{~K}$
(4) $1.254 \times 10^{4} \mathrm{~K}$
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere?
(Given: Mass of oxygen molecule $(\mathrm{m})=2.76 \times 10^{-26} \mathrm{~kg}$ Boltzmann’s constant $\left.k_{B}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$
(1) $5.016 \times 10^{4} \mathrm{~K}$
(2) $8.360 \times 10^{4} \mathrm{~K}$
(3) $2.508 \times 10^{4} \mathrm{~K}$
(4) $1.254 \times 10^{4} \mathrm{~K}$

Answer is (2)

$V_{\text {escane }}=11200 \mathrm{~m} / \mathrm{s}$

Say at temperature $T$ it attains $V_{\text {escape }}$

So, $\sqrt{\frac{3 k_{B} T}{m_{O_{2}}}}=11200 \mathrm{~m} / \mathrm{s}$

On solving we get,

$\mathrm{T}=8.360 \times 10^{4} \mathrm{~K}$