11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
11.23 Wavelength produced by X-ray, 0.45 Å = 0.45 m
Planck's constant, h = 6.626 Js
Speed of light, c = 3 m/s
The maximum energy of a photon is given as:
= = = 4.417 J = eV = 27.6 eV = 27.6 keV
Therefore, the maximum energy of an X-ray photon is 27.6 keV
To get an X-ray of 27.6 keV, the incident
Similar Questions for you
Based on theory
z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV
- (i)
- (ii)
from (i) & (ii)
ev
hu = hu0 + K.E
Cases u = 2u0
h2u0 = hu0 + K.E1
K.E1 = hu0
- (1)
Now, cases 2
h 5u0 = hu0 + k.E2
k.E2 = 4hu0
v2 =
v2 = 2v1
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
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Physics Ncert Solutions Class 12th 2023
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