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K.E. of a subatomic particle is 6.626xx1...

K.E. of a subatomic particle is `6.626xx10^(-19) J`. Calculate its frequency. 2 × 10 − 15 H z 2 × 10 15 H z 2 × 10 − 13 H z 2 × 10 13 H z

A

`2xx10^(-15) Hz`

B

`2xx10^(15) Hz`

C

`2xx10^(-13) Hz`

D

`2xx10^(13) Hz`

Text Solution

Verified by Experts

The correct Answer is:
B
`K.E. =(1)/(2) mv^(2)=(1)/(2) (mv) v=(1)/(2) pv =(1)/(2) (h)/(lamda) xx v =(1)/(2) hv`
frequency, `v=(2KE)/(h) =(6.626xx10^(-19)xx2)/(6.626xx10^(-34)) s^(-1)=2xx10^(15) Hz`
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