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The de - Broglie wavelength of a particl...

The de - Broglie wavelength of a particle accelerated with `150 volt` "potential is" `10^(-10) m`. If it is accelerated by `600 volts p.d`., its wavelength will be

A

`0.25 Å`

B

`0.5 Å`

C

`1.5 Å`

D

`2 Å`

Text Solution

Verified by Experts

The correct Answer is:
B
By using `lambda prop (1)/(sqrt(V)) rArr (lambda_(1))/(lambda_(2)) = sqrt((V_(2))/(V_(1)))`
`rArr (10^(-10))/(lambda_(2)) = sqrt((600)/(150)) = 2 rArr lambda_(2) = 0.5 Å`.
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Knowledge Check

  • The de - Broglie wavelength of a paricle accelerated with 150 volt potential is 10^(-10) m : If it is accelerated by 600 volts p.d. its wavelength will be:

    A
    0.25 Å
    B
    0.5 Å
    C
    1.5 Å
    D
    2 Å

  • The de Broglie wavelength lamda of an electron accelerated through a potential V in volts is

    A
    `(1.227)/(sqrt(V))nm`
    B
    `(0.1227)/(sqrt(V))nm`
    C
    `(0.01227)/(sqrt(V))nm`
    D
    `(12.27)/(sqrt(V))nm`

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

    A
    `(0.287)/(sqrt(V)) Å`
    B
    `(12.27)/(sqrt(V)) Å`
    C
    `(0.101)/(sqrtV)Å`
    D
    `(0.202)/(sqrt(V))Å`

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    Wavelength of an electron accelerated through a potential difference of 1 volt is

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