The wave length of `Kbeta` X-ray of certain metal is `12.42`pm . It takes `10Kev` to remove the electron from M shell of an atom of that metal the minimum accelerating voltage that should be applied across the X-ray tube , so that a `K_(a)` X-ray would be produced is (`hc=1242eVnm`)

Options

A. `10KV`
B. `100KV`
C. `110KV`
D. `90KV`

Answer

C

Explanation

`E_(Kbeta)=(1242eV-nm)/(12.42xx10^(-3nm))=100KeV` and energy of `M` shell is `E_(M) =-10KeV` for K_(alpha) to be produced , electron from k shell is to be knocked out
therefore kinetic energy of the striking electron should be `ge` energy of K shell
`i.e K_(Kbeta) = E_(M)-E_(K) Rightarrow E_(K)=E_(M)-E(Kbeta)`
`i.e e_(K) =-10-100=-110eV`
therefore kinetic energy of striking electron =110eV
`eV=110eV RightarrowV=110V`