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Question

In a Coolidge tube, high speed electrons emitted from the cathode are accelerated through a potential difference of 20kV. Find the minimum wavelength of the X-rays produced.
(Take Planck's constant, h=6.6×1034Js)
  1. 0.36782 A
  2. 0.61875 A
  3. 0.45867 A
  4. 0.72854 A

A
0.36782 A
B
0.61875 A
C
0.72854 A
D
0.45867 A
Solution
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When electrons are accelerated through a potential difference, the work done on them is equal to the change in their kinetic energy. When cathode rays are stopped, they lose kinetic energy which is then converted into X-rays.
Thus, work done on cathode rays (W) =V×q=12mv2 (1)
The energy and frequency of X-rays (any radiation) is related as E=hv. Where h is Planck's constant.
But v=cλ; E=hv=hcλ (2)
Equating (1) and (2) and solving for λ.
hcλ=V×q; λ=hcVq. Substituting h=6.6×1034Js, we have c=3×108ms1.
V=20×103V,q=1.6×1019C
λ=6.6×1034×3×10820×103×1.6×1019=0.61875A

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