X-Rays Numerical Class-12 Nootan ISC Physics Solution Ch-25. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
X-Rays Numerical Class-12 Nootan ISC Physics Solution Ch-25
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-25 | X-Rays |
| Topics | Solved Numericals on X-Rays |
| Academic Session | 2025-2026 |
Solved Numericals on X-Rays
Que-1: If the potential difference applied to a Coolidge tube is 28.8 kV, find out the maximum kinetic energy and velocity of the electrons hitting the target.
Ans- V = 28.8 KV = 28.8 x 10^3 V
Suppose that the maximum kinetic energy is K and velocity is V
K = eV = 1.6 x 10^-19 x 28.8 x 10^3
K = 46.08 x 10^-16 = 4.61 x 10^-15 joule
We know that 1/2 m(Vmax)^2 = 4.61 x 10^-15
1/2 x 9 x 10^-31 x Vmax^2 = 4.61 x 10^-15
Vmax^2 = 4.61 x 2 x 10^-15 / 9 x 10^-31
= 9.22 x 10^16 / 9
= 1.024 x 10^16
Vmax = √(1.024 x 10^16)
= 1.01 x 10^8 m/s
Que-2: 18 kV accelerating voltage is applied across an X-ray tube. Calculate (i) the velocity of the fastest electrons striking the target, (ii) the minimum wavelength in the continuous spectrum of X-rays produced.
Ans- Given eV = 18 KV = 18 x 1.6 x 10^-16 joule
We know that 1/2 m(Vmax)^2 = eV
Vmax^2 = 36 x 1.6 x 10^-16 / 9 x 10^-31 = 4 x 1.6 x 10^15
Vmax^2 = 6.4 x 10^15 = 64 x 10^14
Vmax^2 = √(64 x 10^14) = 8 x 10^7 m/s
We know that
λ = hc/eV = 6.6 x 10^-34 x 3 x 10^8 / 18 x 1.6 x 10^-16
λmin = 0.6875 x 10^10 = 0.6875 Å
Que-3: An X-ray tube is operating at an anode voltage of 20 kV. Calculate the minimum wavelength of X-rays produced. Express your answer in picometre.
Ans- V = 20 kV = 20 x 10^3 V , h = 6.63 x 10^-34 J sec
We know that
λmin = hc/eV
= 6.63 x 10 ^-34 x 3 x 10^8 / 1.6 x 10^19 x 20 x 10^3
= 62.15 x 10^-8
= 62.15 pm
Que-4: Calculate the minimum wavelength of X-rays emitted by an X-ray tube operating at a tube potential of 40 kV. How will this wavelength change if target is made of another metal?
Ans- V = 40 kV = 40 x 10^3 = 4 x 10^4 V
We know that
λmin = hc/eV
= 6.6 x 10 ^-34 x 3 x 10^8 / 1.6 x 10^19 x 4 x 10^4
= 3.1 x 10^-11
= 0.31 Å
Que-5: An X-ray tube is operating at 100 kV. Calculate the maximum frequency of X-rays emitted and maximum velocity of electrons striking the anti-cathode.
Ans- V = 100 kV = 100 x 10^3 = 1 x 10^5 V
We know that
λmin = hc/eV
= 6.6 x 10 ^-34 x 3 x 10^8 / 1.6 x 10^19 x 1 x 10^5
= 0.12375 Å
and Vmax = √(2eV/m)
= √(2 x 1.6 x 10^-19 x 1 x 10^5 / 9 x 10^-31)
= √(3.555 x 10^16)
= 1.88 x 10^8 m/s
Que-6: How much potential difference be applied across an X-ray tube so that the minimum wavelength of the emitted X-rays be 6.6 Å?
Ans- Suppose that the p.d. is V then
λmin = hc/eV
V = hc/eλmin
= 6.6 x 10 ^-34 x 3 x 10^8 / 1.6 x 10^19 x 6.6 x 10^-10
= 1.875 x 10^3
= 1.875 kV
Que-7: Find the maximum frequency of X-rays emitted from an X-ray tube operated at 10 kV. What will be the energy and momentum of the maximum-frequency X-ray photon?
Ans- Suppose that frequency is f , energy E and momentum is p
We know that
hf = eV
f = eV/h
= 1.6 x 10^19 x 10 x 10^3 / 6.6 x 10^-34
= 2.42 x 10^18 Hz
and Energy E = hf = 6.6 x 10^-34 x 2.42 x 10^18
E = 15.972 x 10^-16
= 15.972 x 10^-16 / 1.6 x 10^-19
= 10 x 120^3
= 10 KeV
and momentum p = h/λmin = h/(hc/eV) = eV/c
= 1.6 x 10^19 x 10 x 10^3 / 3 x 10^8
= 5.32 x 10^-24 Kg m/s
Que-8: What element has a Ka X-ray line of wavelength 1.785 Å? R = 109737 cm^-1.
Ans- We know that
1/λ = R(Z – b) for Ka , b = 1
Given λ = 1.785Å = 1.785 x 10^-10
R = 109737 cm^-1 = 109737 1/cm = 109737 x 10^2 m^-1
b = 1.0
∴ 1/1.785 x 10^10 = 109737 x 10^2(Z-1.0)^2[1/1^2 – 1/2^2]
3/4(Z-1)^2 = 10^8/1.785 x 109737
Z-1 = √(678.984)
Z-1 = 26
Z = 27 Mo
–: X-Rays Numerical Class-12 Nootan ISC Physics Solution Ch-25. :–
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