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In an X-ray tube the accelerating voltag...

In an X-ray tube the accelerating voltage is `20 kV`. Two target `A` and `B` are used one by one. For `'A'` the wavelength of the `K_(alpha)` lines is `62 "pm"`. For `'B'` the wavelength of the `L_(alpha)` line is `124 "pm"`. The energy of the `'B'` ion with vacancy in `'M'` shell is `5.5 KeV` higher than the atom of `B`. [ Take `hc=12400 eV Å`]

A

Value of `lambda_(min)` is `0.62Å`

B

`A` will emitte `K_(alpha)`-photon.

C

`B` will emitte `L`-photns.

D

minimum wavelength (in `Å`) of the characterstics X-ray that will be emitted by `'B'` is `0.8Å`.

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
(A) `lambda_(min)=(hc)/(K.E. ofe^(-))=(12400)/(124xx10^(3))Å0.62Å" "62"pm"`
(B) Since `lambda_(min)=62"pm",K_(alpha)` from `A` will not be obtained
(C )`L`- photon can be emitted if electron if electron from `L`-shell can be remove the energy required to removal `L`-shell electrons.
`=55.5 keV+(12400)/(124xx10^(-2))eV=15.5 keV`
As the energy of encoming electron is `20 KeV gt 15.5 keV`, the `L`-shell electron can be removed. Hence `L`, photon can be obtained.
The minimum wavelength will corresponding to the transimition from `oo` to `L`-shell.
`lambda=(12400)/(15.5xx10^(3))A=0.8Å`
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