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51.

(a) Find the maximum frequency of the X-rays emitted by an X-ray tube operating at `30 kV`. (b) An X-ray tube operates at `20 kV`. A particular electron loses `5%` of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon. (c ) An X-ray tube is operated at `20 kV` and the current through the tube is `0.5 mA`. Find (i) the number of electrons hitting the target per second, (ii) the energy falling on the target per second as the kinetic anergy of the electrons and (iii) the cut-off wavelength of the X-rays emitted.

Answer» (a) `lambda_(min)=(hc)/(eV), V=30 kV`
`lambda_(min)=1242/(E(eV))=1242/(30xx10^(3))=41.4xx10^(-3) nm`
`=41.4xx10^(-12)m`
`v_(max)=c/lambda(min)=(3xx10^(8))/(41.4xx10^(-12))=7.2xx10^(18) Hz`
(b) Kinetic energy of electron `=20 keV`
`5%` of this converted to photon
`5/100xx20=1 ke V`
`lambda=1242/(1xx10^(3)) nm=1.242 nm`
(c ) `V=20 kV, i=0.5 mA`
(i) `i=n eimpliesn=i/e=(0.5xx10^(-3))/(1.6xx10^(-19))=3.125xx10^(15)`
(ii) `P=Vi=20xx0.5==10W`
(iii) `lambda_(min)=1242/(20xx10^(3))=0.062 nm`
Discussion
52.

The `K_(alpha)` X-ray emission line of lungsten accurs at `lambda = 0.021 nm`. What is the energy difference between `K`and `L` levels in the atom?A. `0.51 MeV`B. `1.2 MeV`C. `59 keV`D. `13.6 eV`

Answer» Correct Answer - C
`DeltaE=1242/(lambda(nm))eV=1242/0.021=59142 eV~= 59KeV`
Discussion
53.

An x-ray tube is operated at `20 kV` and the current through the tube is `0*5 mA`. Find (a) the number of electrons hitting the target per second, (b) the energy falling on the target per second as the kinetic energy of the electrons and the cutoff wavelength of the X-rays emitted.

Answer» Correct Answer - `10`
Discussion
54.

The wavelength of `K_alpha` and `L_alpha` X-rays of a meterial are 21.3 pm and 141 pm respectively. Find the wavelength of `K_beta` X-ray of the material.

Answer» Correct Answer - `18.5` pm
Discussion
55.

Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `lambda_(0)=(2mc lambda^(2))/h`B. `lambda_(0)=(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`D. `lambda_(0)=lambda`

Answer» Correct Answer - A
`lambda=h/sqrt(2meV) implies V=h^(2)/(2melambda^(2))`
`(hc)/(lambda_(0))=eV=eh^(2)/(2melambda^(2))`
`lambda_(0)=(2mclambda^(2))/h`
Discussion
56.

The `k_alpha` X-rays of aluminium (Z = 13 ) and zinc ( Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley's law `sqrt v = a(Z - b)` to find the wavelength of the `K_alpha` X-ray of iron (Z = 26).

Answer» `sqrt(v)=a(Z-b)` implies `sqrt(c/lambda)=a(Z-b)`
Aluminium: `sqrt(c/887)=a(13-b) ...(i)`
Zinc: `sqrt(c/146)=a(30-b) ...(ii)`
(ii)/(i)
`sqrt(887/146)=(30-b)/(13-b) implies 2.5(13-b)=30-b`
`b=1.4`
`sqrt(c/146)=a(30-b)`
`sqrt((3xx10^(8))/(146xx10^(-12))) =a(30-1.4)`
`1.43xx10^(8)=a(28.6) implies a=5xx10^(7) (Hz)^(1//2)`
Iron: `sqrt(c/lambda)=a(Z-b)=5xx10^(7)(26-1.4)=123xx10^(7)`
`lambda=(3xx10^(8))/((123xx10^(7))^(2))=1.98xx10^(-10) m`
`=198 p m`
Discussion