In a Coolidge tube, the potential difference used to accelerate the electrons is increased from `12.4 kV` to `24.8 kV`. As a result, the difference between the wavelengths of `K_(alpha)`-line and minimum wavelength becomes thrice. The wavelength of the `K_(alpha)` line is `0.25xxKA^(@)`. Find the value of `K`. `(hc)/(e)=(12.4 KVA^(@))`

Explanation

According to question
`3(lambda_(K_(alpha))-(12.4)/(12.4))=lambda_(K_(alpha))-(12.4)/(12.4)`
`rArr lambda_(K_(alpha)=1.25A^(@)`