An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.42Å. What is the maximum energy of photon in the radiation? What is the order of accelerating voltage (for electrons) requirement in such a case? `h=6.62xx10^(-34)Js and e=1.6xx10^(-19)C`

Answer

`29.6 eV, 30 kV`

Explanation

Here, `lambda=0.42 Å=0.42xx10^(-10)m`,
Max. energy of photon,
`E=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(0.42xx10^(-10))J`
`E=(6.63xx3xx10^(-26))/((0.42xx10^(-10))xx1.6xx10^(-19))eV`
`=29.6xx10^(3) eV=29.6 keV`
It shows that the maximum energy of X-ray photon is 29.6 keV. As accelerating voltage provides enerty to the electron for the production of X-rays, so to get the X-rays of energy 29.6 eV, the incident electron must possess the least energy 29.6 keV. therefore the accelerating voltage of the order of 30 kV is required for producing X-rays.