RAJASTHAN PMT Rajasthan - PMT Solved Paper-2004

  • question_answer
    Thermal speed of free electrons is of the order of :

    A)  \[{{10}^{-5\,}}\,m/s\]                   

    B)         \[{{10}^{2}}\,m/s\]

    C)  \[{{10}^{8}}\,m/s\]        

    D)         \[{{10}^{5}}\,m/s\]

    Correct Answer: D

    Solution :

    Thermal energy of free electron is \[=\frac{3}{2}kT\] Let \[\upsilon \] be the thermal speed of electron, then \[\frac{1}{2}m{{\upsilon }^{2}}=\frac{3}{2}kT\]             \[\upsilon =\sqrt{\frac{3kT}{m}}\] \[\Rightarrow \]\[\upsilon =\sqrt{\left[ \frac{3\times 1.38\times {{10}^{-23}}\times 300}{9.1\times {{10}^{-31}}} \right]}\]                      \[={{10}^{5}}\,m/s\]


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