Photoelectric EffectHard
Question
In an x-ray tube the voltage applied is 20KV. The energy required to remove an electron from L shell is 19.9 KeV. In the x-rays emitted by the tube (hc = 12420 ev(Ao))
Options
A.minimum wavelength will be 62.1 pm
B.energy of the characterstic x-rays will be equal to or less than 19.9 KeV
C.Lαx-ray may be emitted
D.Lα x-ray will have energy 19.9 KeV
Solution
(A) Minimum wavelength will correspond to maximum energy i.e.,from ∞ to k.
ᐃE = 19.9 KeV
∴ λmin =
= 0.62 (Ao)
= 62 pm.
(B) Energy of the characteristic x-rays will be less than corresponding to ∞ to k-shell, hence than 19.9 KeV.
ᐃE = 19.9 KeV
∴ λmin =
= 0.62 (Ao)= 62 pm.
(B) Energy of the characteristic x-rays will be less than corresponding to ∞ to k-shell, hence than 19.9 KeV.
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