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Planck's constant (h) and de Broglie wav...

Planck's constant (h) and de Broglie wavelength `(lambda)` are related through the equation `h=lambdasqrt(2mE)`, where 'm' and 'E' denote the mass and kinetic energy respectively of the moving particle. The dimensional formula of h is given by

A

`[M^(2)L^(2)T^(-2)]`

B

`[M^(1)L^(2)T^(-1)]`

C

`[M^(1)L^(1)T^(-1)]`

D

`[M^(2)L^(-1)T^(-2)]`

Text Solution

Verified by Experts

The correct Answer is:
b
`h=lambdasqrt(2mE)" but "[E]=[M^(1)L^(2)T^(-2)]and[lambda]=[L^(1)]`
`therefore [h]=[L^(1)][M^(1)M^(1)L^(2)T^(-2)]^(1//2)=[M^(1)L^(2)T^(-1)]`
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Knowledge Check

  • The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as

    A
    `lambda prop E`
    B
    `lambda prop sqrt(E)`
    C
    `lambda prop 1//E`
    D
    `lambda prop 1//sqrt(E)`

  • If the kinetic energy of a moving particle is E , then the de-Broglie wavelength is

    A
    `lamda=hsqrt(2mE)`
    B
    `lamda=sqrt((2mE)/h)`
    C
    `lamda=h/(sqrt(2mE))`
    D
    `lamda=(hE)/(sqrt(2mE))`

  • The de - Broglie wavelength lambda associated with an electron having kinetic energy E is given by the expression

    A
    `(h)/(sqrt( 2 m E))`
    B
    `( 2h)/(m E)`
    C
    `2 mhE`
    D
    `( 2 sqrt(2 mE))/(h)`

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