One increasing the operating voltage in a x-ray tube by `1.5` times, the shortest wavelength decreases by `26 "pm"`. Find the original value of operating voltage.

Answer

`v_(1)=(hc)/(1.5(lambda_(1)+26xx10^(-12)))V=15.9 kV`

Explanation

`KE=(hc)/(lambda_(min))`
`lambda_(min)=(hc)/(KE)`
`lambda_(1)-lambda_(2)=1.24xx10^(4)((1)/(v)-(1)/(1.5v))Å`
`26xx10^(-2)=1.24xx10^(4)xx(1)/(3V)`
`:. V=(1.24xx10^(4))/(3xx26xx10^(-2))=15.9xx10^(3)V`