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

The wave length of `Kbeta` X-ray of certain metal is `12.42`pm . It ta

1.	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`)A. `10KV`B. `100KV`C. `110KV`D. `90KV`
Answer» Correct Answer - C `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`

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