Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q e =1.60*10^

Question

Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q e =1.60*10^ -19 kg and m e =9.109*10^ -31 kg.)

Please please helpppp?

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Edana Edana 5 years 2021-08-02T20:38:49+00:00 1 Answers 38 views 0

Answers ( )

  1. Helga
    0
    2021-08-02T20:39:57+00:00
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    Answer:

    \lambda=1.37\times 10^{-11}\ m

    Explanation:

    Give that,

    The potential difference of the electrons, V=8\times 10^{3}\ V

    We need to find the wavelength of the electrons.

    Using the conservation of energy,

    2meV=\dfrac{h^2}{\lambda^2}\\\\\lambda=\sqrt{\dfrac{h^2}{2meV}}

    Put all the values,

    \lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m

    So, the wavelength of the electrons is 1.37\times 10^{-11}\ m.

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