X-ray are emitted by a tube containing the electron gt Noibum `(Z = 41)` as anticathode and the `K_(alpha)` X-ray an allowed to be incident on an unknown gas containing hydrogen- like atom ions. It is found that the X-ray cause the emission of photoelectrons with an energy of `2.7 keV` from these ions. Find
a. the minimum voltage at which the X-ray tube should be opreated so that the momentum of the emitted photoelectrons is doubled.
b. the approximate value of `Z` for the target of the anticathode after the momentum has been doubled.
c. the approximate value of `Z` the atomic number of the atom of the gas.
d. the number of the ion produced in the gas caused by X-ray , if the inetensity of X-ray is `100 mWm^(-2)` and `1%` of the X-rays cause ionization.

Answer

`24.5 keV, = 850 per m^(3)` b. 50 c. 32

Explanation

The wavelength of the `K_(alpha)` radiation from `Nb` is
`lambda = (4 hc)/(3 R) (1)/((Z - 1)^(2)) = (4 hc)/(3 R) (1)/(40^(2))`
`E_(K_alpha) = 1.63 keV`
`Z_(gas)^(2) = 1000 implies Z_(gas) = 32`
If the momentum of the electron is doubled, then the `KE` is increased by `4` times
Therefore , the minimum energy of the X-rays is `(16.3 + 3 xx 2.7) keV = 24.5 keV`.
Since `E_(K_alpha) prop (Z - 1)^(2)`
`((Z - 1)/(40))^(2) = ((24.5)/(16.3)) implies Z = 50`
The voltage at which the tube should be operated should be greater than `24.5 kV`.
The number of X-ray photons incident`= 100 mWm^(-2)`
`implies l = (100 xx 10^(-3) xx 10^(19))/(1.6 xx 24.5 xx 10^(3)) implies m^(-2)`
Number of ions produced in `1 s (n) = 1% of(l)/(C )`
`(100 xx 10^(-3) xx 10^(-2) xx 10^(19))/(1.6 xx 24.5 xx 10^(3) xx 3 xx 10^(8)) m^(-3) = 850 per m^(-3)`