The photoelectric threshold wavelength of silver is `3250 xx10^-10 m.` The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx10^-10` m is
(Given: `h = 4.14 xx10^-15 eVs and c = 3xx10^8 ms^-1)`

Options

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

Answer

A, B

Explanation

`(hc)/(lamda) - (hc)/(lamda_(th)) = 1/2 xxm_e xxv^2`
`v= sqrt((2hc)/(m_e)(1/lamda-1/lamda_(th)))`
`v= sqrt((2xx12400xx 1.6 xx10^(-19))/(9.1 xx10^(-31))[1/(2536)-1/(3250)])`
Solving this we get
`v = 6 xx10^5 ms^(-1)`