The photoelectric threshold wavelength of silver is `3250 xx 10^(-10) m`. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx 10^(-10) m` is
`(Given h = 4.14 xx 10^(6) ms^(-1) eVs` and `c = 3 xx 10^(8) ms^(-1))`

Options

A. `~~0.6xx10^(6) ms^(-1)`
B. `~~61xx10^(3) ms^(-1)`
C. `~~0.3xx10^(6) ms^(-1)`
D. `~~6xx10^(5) ms^(-1)`

Answer

A, D

Explanation

`phi=1242/325=3.82 eV`
`E=1242/253.6=4.89 eV`
`K_(max)=E-phi=1.077 eV`
`1/2mv_(max)^(2)=1.077xx1.6xx10^(-19)`
`v_(max)=((2xx1.077xx1.6xx10^(-19))/(9.1xx10^(-31)))^(1//2)=0.6xx10^(6)m//s`