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An electron is accelerated under a poten...

An electron is accelerated under a potential difference of 64 V, the de-Brogile wavelength associated with electron is `[e = -1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31)kg, h = 6.623 xx 10^(-34) Js]`

A

`1. 53 Å`

B

`2. 53 Å`

C

`3. 35 Å`

D

`4. 54 Å`

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A
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Calculate the wavelength of an electron that has been accelerated in a particle acceleratior through a potential difference of 100 million volts. (1 eV = 1.6 xx 10^(-19)C, m_e = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) Js) .

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potentiation difference of 100 million volts [ 1eV = 1.6 xx 10^(-19)J, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s, c = 3.0 xx 10^(8) m s^(-1)]

Knowledge Check

  • Photon having energy equivalent to the binding energy of 4th state of He^(+) ion is used to eject an electron from the metal with KE/2eV . If electron is further accelerated through a potential difference of 4V then the minimum value of de Broglie wavelength associated with the electron is : (h = 6.6 xx 10^(-34) J-s , m_(e) = 9.1 xx 10^(-31) kg . 1 eV = 1.6 xx 10^(-19) J )

    A
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    `11 "Å"`

  • An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m = 9 xx 10^(-33) kg)

    A
    `12.2 xx 10^(-13)`m
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  • The kinetic energy of an electron is 5 eV . Calculate the de - Broglie wavelength associated with it ( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)

    A
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    `109 Å`
    C
    `2.7 Å`
    D
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