The wavelength of `K_(alpha)` line for copper is `1.36^(@)`. Calculate the ionisation potential of a K shell electron in copper.

Answer

`1.22xx10^(4)V`

Explanation

form the relation `1/lambda=RZ^(2)(1/(n_(1)^(2))-1/(n_(2)^(2)))`
for `K_(alpha)` line, `n_(1)=1 , n_(2)=2`
`:. 1/lambda=RZ^(2)(1/(1^(2))-1/(2^(2)))=3/4RZ^(2)`
`:. RZ^(2)=4/(3lambda)........(i)`
Ionisation energy of K shell electron is
`E=(2pi^(2)mK^(2)Z^(2)e^(4))/(h^(2)) (1/(1^(2))-1/(oo)^(2))`
`=(2pi^(2)mK^(2)Z^(2)e^(4))/(h^(2)(ch)) (ch)=RZ^(2)(ch)`
Using (i),
`E=4/(3lambda)ch=(4xx3xx10^(8)xx6.63xx10^(-34))/(3xx1.38xx10^(-10))`
`=19.5xx10^(-16)J`
`E=(19.5xx10^(-16))/(1.6xx10^(-19))eV=1.22xx10^(4)eV`
`:.` Ionisation potential `=1.22xx10^(4)"volt"`