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The de Broglie wavelenght (lambda) assoc...

The de Broglie wavelenght `(lambda)` associated with a photoelectron varies with the frequency `(v)` of the incident radiation as,[`v_(0)` is threshold frequency]:

A

`lamda prop(1)/((v-v_(0))^(3/2))`

B

`lamda prop(1)/((v-v_(0))^(1/4))`

C

`lamda prop(1)/((v-v_(0))^(1/2))`

D

`lamda prop (1)/((v-v_(0)))`

Text Solution

Verified by Experts

The correct Answer is:
C
`E = hu`
`(hu-hr_(0))=K.E.` (from photoelectric effect)
`lamda = (h)/(sqrt(2mK.E. )), " "lamda prop (h)/(sqrt((u -u _(0))))implies lamda prop (1)/((u-u_(0))^(1//2))`
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