If the photon of the wavelength `150 p m` strikes an atom and one of its inner bound electrons is ejected out with a velocity of `1.5xx10^(7) ms^(-1)`, calculate the energy with which it is bound to the nucleus.

Options

A. `1.2 xx 10^(2) eV `
B. `2.15 xx 10^(3)` eV
C. `7.6 xx 10^(3)` eV
D. `8.12 xx 10^(3)` eV

Answer

C

Explanation

Total energy (E) of any photon is given by the relation :
`E= (hc)/(lambda)`
where, h = Planck’s constant `= 6.6 xx 10^(-34) J-s`
c = velocity of light = `3xx 10^(-10)` m/s
`lambda` = wave -length `= 1.5 xx 10^(-10) m ` i.e , (150 pm )
Thus `E = (hc)/(lambda)= (6.6xx10^(-34) xx 3 xx 10^(8))/(1.5 xx 10^(-10)), E = 1.32 xx 10^(-15)` J
and, energy of ejected electron ( E )
`E= (1)/(2) mv^(2)`
where
m = mass of electron `= 9.1 xx 10^(-31) ` kg
v = velocity of electron `= 1.5 xx 10^(7)` m/s
`E = (1)/(2) mv^(2 ) = (1)/(2) xx 9.1 xx10^(-31) xx (1.5 xx 10^(7))^(2)`
`E . = 1.024 xx 10^(-16)`
Thus, total energy of photon = binding energy of electron (B) + energy of ejected electron (E )
Thus `1.32 xx 10^(-15) = (B) +(E)`
`:. (B) =( E) - (E ) = [ 1.32xx 10^(-15)] - [1.024 xx 10^(-16)]`
`= 1.2176 xx 10^(-15) J`
`=(1.2176 xx 10^(-15))/(1.6 xx10^(-19)) eV = 7.6 xx10^(3) ` eV