Question

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping form the Earth's atmosphere? (Given : mass of oxygen molecule $\left(m\right)=2.76\times {10}^{-26}$ kg, Boltzmann constant $(K_B=2.76\times {10}^{-23}{JK}^{-1})$

• A. $2.508\times {10}^{4}K$
• B. $8.360\times {10}^{4}K$
• C. $5.016\times {10}^{4}K$
• D. $1.254\times {10}^{4}K$

Answer 1

The correct option is B $8.360\times {10}^{4}K$

At the minimum velocity with which the body must be projected vertically upwards so that it could escape from the earth's atmosphere is its escape velocity $V_e$

As

$v_l=\sqrt{2gR}$

Substituting the value of g $\left(9.8m{s}^{-2}\right)$ and radius of earth $(R=6.4\times {10}^{6}m)$ We get

$v_e=\sqrt{2\times 9.8\times 64\times {10}^6}$

$\cong 11.2km{s}^{-1}=11200m{s}^{-1}$

Let the temperature of molecule be $T$ when it attains $v_e$.

According to the question,

$v_{rms}=v_e$

where $v_{rms}$ is the $rms$ speed of the oxygen molecule,

$\Rightarrow \sqrt{\frac{3k_{B}T}{m{o}_2}}=11.2\times {10}^3$

or $T=\frac{(11.2\times {10}^3{)}^2(m{o}_2)}{\left(3k_B\right)}$

Substituting the given values i.e

$R_B=1.38\times {10}^{-23}J{K}^{-1}$ and

$m{o}_2=m=2.76\times {10}^{-26}kg$

We get,

$T=\frac{(11.2\times {10}^3{)}^2(2.76\times {10}^{-26})}{(3\times 1.38\times {10}^{-23})}$

$=8.3626\times {10}^{4}k$