JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    The de-Broglie wavelength of helium atom at room temperature is

    A) \[6.6\times {{10}^{-34}}m\]                        

    B) \[4.39\times {{10}^{-10}}m\]

    C) \[7.34\times {{10}^{-11}}m\]                     

    D)  \[2.335\times {{10}^{-20}}m\]

    Correct Answer: C

    Solution :

    \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 298}{4\times {{10}^{-3}}}}=1363\,\,m{{s}^{-1}}\]    \[\lambda =\frac{h}{mv}=\frac{6.626\times {{10}^{-34}}\times 6.023\times {{10}^{23}}}{4\times {{10}^{-3}}\times 1363}\]       \[=7.32\times {{10}^{-11}}m\]


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