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

Answer

A, D

Explanation

The maximum kinetic energy is given as
`K_(max)=hupsilon-phi_0=hupsilon-hupsilon_0=(hc)/lamda-(hc)/lamda_0`
where `lamda_0` = threshold wavelength or `1/2mv^(2) =(hc)/lamda =(hc)/lamda_0`
Here. `h = 4.14 xx 10^(-15) eV s,C = 3 xx 10^(8) ms^(-1)`
`lamda_0 = 3250 xx 10^(-10) m = 3250 Å`
`lamda = 2536 xx 10^(-10) m = 2536 Å , = 9.1 xx 10^(-31) kg`
`hc = 4.14 xx 10^(-15) eVs xx 3 xx 10^(8) ms^(-1) = 12420 eV Å`
`:. 1/2 mv^2 = 1240 [1/(2536)-1/(3250)] = eV = 1.076 eV` `v^2 =(2.152 eV)/m = (2.152xx1.6xx10^(-19))/(9.1 xx10^(-31))`
`:. v ~~ 6 xx 10^(5) ms^(-1) = 0.6 xx 10^(6) ms^(-1)`
Note : Options (a) and (d) are same . so both are correct.