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Light of two different frequencies, whose photons have energies 2eV and 5eV respectively, successively illuminates a metal of work function 1.0 eV. The ratio of maximum kinetic energy of the emitted photoelectrons will be

A

`2:5`

B

`1:5`

C

`1:4`

D

`1:2`

Text Solution

Verified by Experts

The correct Answer is:
C
As per relation `K_(max)=E-phi_(0)`
`(K_(1))/(K_(2))=(E_(1)-phi_(0))/(E_(2)-phi_(0))=((2-1)eV)/((5-1)eV)1:4`.
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