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Question: Answered & Verified by Expert

In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24.8 kV to 49.6 kV. As a result, the difference between the wavelength of Kα-line and minimum wavelength becomes two times. The initial wavelength of the Kα-line is [Take hce=12.4 kV A°]

PhysicsAtomic PhysicsJEE Main
Options:
  • A 32A°
  • B 34A°
  • C 52A°
  • D 54A°
Solution:
1363 Upvotes Verified Answer
The correct answer is: 34A°

Let ∆E be the energy of kα-line then

hcλKα=ΔEKα
λKα=hcΔEKα
Now the cutoff wavelengths in the two cases are

 λ1min=hce×24.8 kV=12.424.8A°=12A° 

λ2min=hce×49.6 kV=12.449.6A°=14A°
2λkα-12=λkα-14
2λkα-1=λkα-14
λkα=1-14
λkα=34Å

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