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de-Broglie wavelength associated with an electron accelerated through a potential difference V is `lamda`. What will be its wavelength when the accelerating potential is increased to 4 V?

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As `lamdaprop(1)/(sqrtV)`, hence on increasing accelerating potential from V to 4V, de-Broglie wavelength decreases from `lamda` to `(lamda)/(2)`.
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Knowledge Check

  • The de-Broglie waves associated with an electron accelerated through a potential difference of 121V is :

    A
    1.227 nm
    B
    12.270 nm
    C
    0.112 nm
    D
    11.200 nm

  • The wavelength associated with an electron accelerated through a potential difference of 100 V is nearly

    A
    `100 Å`
    B
    `123 Å`
    C
    `1.23 Å`
    D
    `0.123 Å`

  • What is de-Broglie wavelength of the electron accelerated through a potential difference of 100V?

    A
    0.12 Å
    B
    12 Å
    C
    1.22 Å
    D
    None of these

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