In a photoelectric experiment, with light of wavelength `lambda`, the fastest election has speed v. If the exciting wavelength is changed to `(3lambda)/4`, the speed of the fastest emitted electron will become

Options

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

Answer

A

Explanation

Let `E=(hc)/lamda`
`E.=(hc)/(3lamda/4)=4/3E`
The work function for both the cases is same.
`E=phi+1/2mv^2" "...(1)`
`E.=phi+1/2mv^2`
`(4E)/3=phi+1/2mv^2`
Subtracting equation (1), we get
`E/3=1/2mv^2-1/2mv^2`
`1/3(phi+1/2mv^2)=1/2mv^2-1/2 mv^2`
`phi/3+1/6mv^2+1/2mv^2=1/2mv.2`
`(2phi+4mv^2)/(3m)=v.2`
`(2phi)/(3m)+4/3v^2=v^2`
`impliesv.gtv(4/3)^(1//2)`