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When light of wavelength 300 nm falls on...

When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitted , light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

A

`1:2`

B

`2:1`

C

`4:1`

D

`1:4`

Text Solution

Verified by Experts

The correct Answer is:
B
`phi_0 prop 1/(lamda_0) implies (phi_(0_1))/(phi_(0_2))=(lamda_(0_2))/(lamda_(0_1))`
`implies(phi_(0_1))/(phi_(0_2))=2/1 `
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