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^(6)ms^(-1)`
C. `=61xx10^(3)ms^(-1)`
D. `=0.3xx10^(6)ms^(-1)`

Answer

A, B

Explanation

Thinking Process Applying Einstein's photoelectric equation, Kinetic energy of emitted electron can be given vby
`K=(1)/(2)mv^(2)=hv-hv_(0)=(hc)/(gamma)-(hc)/(gamma_(0))`
Given threshold wavelength, `gamma_(0)=3250xx10^(-10)` m Wavelength of utraviolet light, `gamma =2536xx10^(-10) m` Let, velocity of rejected electron be v,
Now applying Einsten's photolectric equation we have
" " `E=K+phi_(0)rArrhv=(1)/(2)m_(e)V^(2)+hv_(0)`
`rArr" "(1)/(2)m_(e)V^(2)=hv-hv_(0)=hc((1)/(gamma)-(1)/(gamma_(0)))`
`rArr Velocity of electron V= sqrt(2hc)/(m_(e))((1)/(gamma)-(1)/(gamma_(0)))`
`=sqrt(2xx4.14xx10^(-15)xx1.6xx10^(-19)1.6xx10^(-19)xx3xx10^(9))/(9.1xx10^(-31))((3250-2536)/(3250xx2536))`
`0.6xx106(6) ms^(-1)=6xx10^(5)ms^(-6)`