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A alpha -parhticle moves in a circular ...

A `alpha` -parhticle moves in a circular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. The de-Broglie wavelength assocaiated with the particle will be

A

`1 Å`

B

` 0.1 Å`

C

`10 Å`

D

` 0.01 Å`

Text Solution

Verified by Experts

The correct Answer is:
D
`lamda =h/P`
Since ,` r = (mv)/(qB)`
`implies mv = qrB = 2e xx0.83 xx10^(-2) xx1/4`
`lamda = (6.6 xx10^(-34) xx4)/(2 xx1.6 xx10^(-19)xx0.83 xx10^(-2))`
`implies lamda = 0.01 Å`
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