JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    If the de Broglie wavelength of an electron is equal to \[{{10}^{-3}}\] times the wavelength of a photon of frequency \[6\times {{10}^{14}}Hz.\]then the speed of electron is equal to
    (Speed of light\[=3\times {{10}^{8}}m/s\]
    Planck?s constant \[=6.63\times {{10}^{-34}}Js\]
    Mass of electron\[=9.1\times {{10}^{-31}}kg\])
      [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[1.7\times {{10}^{6}}m/s\]                             

    B) \[1.8\times {{10}^{6}}m/s\]

    C) \[1.45\times {{10}^{6}}m/s\]

    D)                  \[1.1\times {{10}^{6}}m/s\]

    Correct Answer: C

    Solution :

    The wavelength of photon of frequency v is \[\lambda =\frac{c}{\upsilon }\]                                                  ?(i) The de Broglie wavelength of the electron is \[{{\lambda }_{e}}=\frac{h}{mv}\]      \[\Rightarrow \]\[v=\frac{h}{m{{\lambda }_{e}}}=\frac{h\upsilon }{mc\times {{10}^{-3}}}\]                    [Given\[{{\lambda }_{e}}={{10}^{-3}}\lambda \]] \[v=\frac{6.63\times {{10}^{-34}}\times 6\times {{10}^{14}}}{9.1\times {{10}^{-31}}\times 3\times {{10}^{8}}\times {{10}^{-3}}}\]\[=1.46\times {{10}^{6}}m/s\]


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