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For a certain metal v is the five times ...

For a certain metal v is the five times of `v_(0)` and the maximum velocity of coming out photons is `8 xx 10^(6)m//s`. If `v = 2v_(0)`, then maximum velocity of photoelectrons will be

A

`4 xx 10^(6)m//s`

B

`6xx10^(6)m//s`

C

`2xx10^(6)m//s`

D

`1xx10^(6)m//s`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(2)m(8xx10^(6))^(2)=h(5v_(0)-v_(0))" "...(i)`
`" and "(1)/(2)mv^(2)=h(2v_(0)-v_(0))" "...(ii)`
Divding Eq. (ii) by Eq. (i), we get
`((8xx10^(6))^(2))/(v^(2))=(4v_(0))/(v_(0))impliesv_(2)=((8xx10^(6))^(2))/(4)`
`v=(8xx10^(6))/(2)impliesv=4xx10^(6 )m//s`
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