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AMU Medical AMU Solved Paper-2005

  • question_answer
    At what temperature, hydrogen molecules will escape from the earths surface ? (Take mass of hydrogen molecule \[=\text{ }0.\text{34}\times \text{1}{{0}^{-26}}\text{ kg},\] Boltz-mann constant \[=\text{1}.\text{38}\times \text{l}{{0}^{-23}}\text{ J}/\text{K},\] radius of earth \[=\text{ 6}.\text{4}\times \text{1}{{0}^{\text{6}}}\text{ m}\]and acceleration due to gravity \[=\text{9}.\text{8 m}/{{\text{s}}^{\text{2}}})\]

    A)  10 K                                      

    B)  102 K

    C)  103 K                                   

    D)  104 K

    Correct Answer: D

    Solution :

                     The root mean square velocity of gas molecules is equal to                 \[{{v}_{rms}}=\sqrt{\frac{3\,kT}{M}}\]                   ?. (i) where T is temperature, and M the molecular weight. Also, escape velocity                 \[{{v}_{e}}=\sqrt{2g{{R}_{e}}}\]                ?.. (ii) where \[{{R}_{e}}\] is radius of earth. Given.   \[{{v}_{rms}}={{v}_{e}}\]                 \[\sqrt{\frac{3kT}{M}}=\sqrt{2g\,{{R}_{e}}}\] \[\Rightarrow \]               \[T=\frac{2g\,{{R}_{e}}M}{3\,k}\] Given,   \[{{R}_{e}}=6.4\times {{10}^{6}}m\],       \[g=9.8\,m/{{s}^{2}}\],                 \[M=0.34\times {{10}^{-26}}kg\]                 \[k=1.38\times {{10}^{-23}}J/K\]. \[\Rightarrow \]               \[T=\frac{2\times 9.8\times 6.4\times {{10}^{6}}\times 0.34\times {{10}^{-26}}}{3\,(1.38\times {{10}^{-23}})}\] \[\Rightarrow \]               \[T={{10}^{4}}K\] Therefore, 104 K is the temperature at which hydrogen molecules will escape from earths surface.


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