The energy of a tungsten atom with a vacancy in L small is `11.3 KaV`. Wavelength of `K_(alpha)` photon for tungsten is `21.3 "pm"`. If a potential difference of `62 kV` is applied across the `X`-raystube following characterstic X-rays will be produced.

Options

A. K,L series
B. only `K_(alpha)` & `L` series
C. only L series
D. None of these

Answer

C

Explanation

`DeltaE = (hc)/(lambda) = (1240)/(lambda(nm)) rArr = (1240)/(21.3 xx 10^(-3))`
`E_(2) = 58.21 KeV , DeltaE_(1) = 11.3 KeV, DeltaE = 69.51 KeV`
`DeltaE lt 62 KeV` Therefore only `1`, series will be produced.