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. `~~6xx10^(5)ms^(-1)`
B. `~~0.6xx10^(8)ms^(-1)`
C. `~~61xx10^(3)ms^(-1)`
D. `~~0.3xx10^(6)ms^(-1)`

Answer

d

Explanation

Here, `lambda_(0)=3250xx10^(-10)m`,
`lambda=2536xx10^(-10)m`
`h=4.14xx10^(-15)eVs`
`=4.14xx10^(-15)xx1.6xx10^(-19)Js`
`1/2 mv^(2)=(hc)/(lambda)-(hc)/(lambda_(0))=hc[(lambda_(0)-lambda)/(lambda_(0)lambda)]`
`=((4.14xxx10^(-15)xx1.6xx10^(-19))xx(3xx10^(8))xx(3250-2536)xx10^(-10))/((3250xx2536)xx10^(-20))`
`=1.077xx1.6xx10^(-19)`
`v=sqrt((2xx1.077xx1.6xx10^(-19))/(9.1xx10^(-31)))`
`=0.6xx10^(6)m//s=6xx10^(5)m//s`