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A photon having 8 eV energy is incident ...

A photon having 8 eV energy is incident on metal suraface with threshold frequency of `1.6xx10^(15)` Hz .Maximum kinetic energy of photoelectron emitted will be…….
`h=6.6xx10^(-34)JS,c=3xx10^(8)m//s`
`eV=1.6xx10^(-19)JS`

A

1.4 eV

B

0.4 eV

C

4.2 eV

D

2.8 eV

Text Solution

Verified by Experts

The correct Answer is:
A

`(1)/(2)mv_(max)^(2)=E-phi_(0)`
`=(E-(hf_(0))/(e))eV [because phi_(0)=(hf_(0))/(e)]`
`=(8-(6.6xx10^(-34)xx1.6xx10^(15))/(1.6xx10^(-19)))`
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