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The work function of caesium is 2.14 eV....

The work function of caesium is 2.14 eV.Find (a) the threshold frequency for caesium,and (b)The wavelength of the incident light if the photocurrent is brought to zero by stopping potential of 0.60 V.

A

`v_(0)=4.54xx10^(14)Hz,lambda=454nm`

B

`v_(0)=5.16xx10^(14)Hz,lambda=516nm`

C

`v_(0)=5.16xx10^(14)Hz,lambda=454nm`

D

`V_(0)5.16xx10^(14)Hz,lambda=414 nm`

Text Solution

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The correct Answer is:
C
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