Radiation of wavelength `lambda` is incident on a photocell.The fastest emitted electron has
speed v.If wavelength is changed to `(3lambda)/(4)`
the speed of the fastest emitted electron will be

Options

A. `gtv((4)/(3))^((1)/(2))`
B. `ltv((4)/(3))^((1)/(2))`
C. `=v((4)/(3))^((1)/(2))`
D. `=v((3)/(4))^((1)/(2))`

Answer

A

Explanation

`K_(max)=hf-phi_(0)`
`=(hc)/(lambda)-phi_(0)`
`therefore (hc)/(lambda)=K_(max)+phi_(0)`
For first case,
`(hc)/(lambda)=(1)/(2)mv^(2)+phi_(0)`
For firat case,
`(hc)/(lambda)=(1)/(2)mv^(2)+phi_(0)` ……(1)
For second case,
`(1)/(2)mv_(1)^(2)=(hc)/(lambda_(1))-phi_(0)`
`(1)/(2)mv_(1)^(2)=(4hc)/(3lambda)-phi_(0)[because lambda-(1)=(3lambda)/(4)]`
`therefore (1)/(2)mv_(1)^(2)=(4)/(3)((1)/(2)mv^(2)+phi_(0))-phi_(0)` [ `because` From result (1)]
`therefore (1)/(2)mv_(1)^(2)=(2)/(3)mv^(2)+(4)/(3)phi_(0)-phi_(0)`
`therefore (1)/(2)mv_(1)^(2)=(2)/(3)mv^(2)+(phi_(0))/(3)`
`therefore v_(1)^(2)=(4)/(3)v^(2)+(2)/(3)(phi_(0))/(m)`
`therefore v_(1)=sqrt((4)/(3)v^(2))+sqrt((2)/(3)(phi_(0))/(m)) (2)/(3)[(phi_(0))/(m)="constant"]`
`therefore v_(1)sqrt((4)/(3))v+"constant" therefore gt v((4)/(3))^((1)/(2))`