A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelen

Question

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 750 nm. What is the potential difference though which this electron was accelerated? (h = 6.626 × 10-34 J · s, e = – 1.60 × 10-19 C, mel = 9.11 × 10-31 kg)

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RobertKer 1 year 2021-08-31T14:41:32+00:00 1 Answers 16 views 0

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  1. thnahdat
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    2021-08-31T14:42:59+00:00
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    Answer:

    Potential difference though which the electron was accelerated is [tex]2.67\times 10^{-6}\ uV\ .[/tex]

    Explanation:

    Given :

    De Broglie wavelength , [tex]\lambda=750\ nm.[/tex]

    Plank’s constant , [tex]h=6.626\times 10^{-34}\ J.s \ .[/tex]

    Charge of electron , [tex]e=-1.6\times 10^{-19}\ C.[/tex]

    Mass of electron , m=9.11\times 10^{-31}\ kg.[tex]m=9.11\times 10^{-31}\ kg.[/tex]

    We know , according to de broglie equation :

    [tex]\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .[/tex]

    Now , we know potential energy applied on electron will be equal to its kinetic energy .

    Therefore ,

    [tex]qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}[/tex]

    Putting all values in above equation we get ,

    [tex]V=2.67\times 10^{-6}\ uV .[/tex]

    Hence , this is the required solution.

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