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An electron with initial kinetic energy ...

An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron becomes

A

`1 Å`

B

`sqrt(1.5Å)`

C

`sqrt(3)Å`

D

`12.27Å`

Text Solution

Verified by Experts

The correct Answer is:
A
For electron `lambda_(db)=sqrt(150/(100+50))=1 Å`
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Knowledge Check

  • Kinetic energy of an electron accelerated in a potential difference of 100 V is

    A
    `1.6 xx 10^(-17) J`
    B
    `1.6 xx 10^(21) J`
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    `1.6 xx 10^(-29) J`
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  • The kinetic energy of an electron accelerated from rest through a potential difference of 5V will be

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    `5J`
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    `5erg`
    C
    `5eV`
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    `8 xx 10^(-10)eV`

  • An alpha - particle has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 150 volt . If a photon has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 25 volt , then find the approximate ratio of the final wavelengths associated with the proton and the alpha -particle .

    A
    5
    B
    4
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    D
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