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NEET Sample Paper NEET Sample Test Paper-49

  • question_answer
    An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron is \[{{E}_{e}}\] and \[{{P}_{e}}\] and that of photon is \[{{E}_{ph}}\] and \[{{P}_{ph}}\] respectively, then correct statement is-

    A) \[\frac{{{E}_{e}}}{{{E}_{ph}}}\,\,=\,\,\frac{2c}{v}\]               

    B) \[\frac{{{E}_{e}}}{{{E}_{ph}}}\,\,=\,\,\frac{v}{2c}\]

    C) \[\frac{{{P}_{e}}}{{{P}_{ph}}}\,\,=\,\,\frac{2c}{v}\]               

    D) \[\frac{{{P}_{e}}}{{{P}_{ph}}}\,\,=\,\,\frac{v}{2c}\]

    Correct Answer: B

    Solution :

    \[\frac{{{E}_{e}}}{{{E}_{p}}}=\frac{\frac{1}{2}m{{v}^{2}}}{h\upsilon }=\frac{1}{2}v\times \frac{mv}{h}.\frac{h}{h\upsilon }\] But \[\frac{h}{mv}\,\,=\,\,\frac{h}{h\upsilon }\] \[\therefore \,\,\frac{{{E}_{e}}}{{{E}_{ph}}}=\frac{1}{2}v\,\left[ \frac{mv}{h}.\frac{h}{h\upsilon }.\frac{c}{c} \right]=\frac{v}{2c}\]


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