HC Verma Concepts Of Physics Exercise Solutions Of Chapter 44 (X-Rays)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 44 - X-Rays:

EXERCISE

Planck constant, h = 4.14 * 10^–15 eV-s or, 6.63 * 10^–34 J-s; speed of light, c = 3 * 10⁸ m/s.

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1. Find the energy, the frequency and the momentum of an X-ray photon of wavelength 0.10 nm.

Sol:

Given: wavelength, λ = 0.10 nm = 1 * 10^–10 m.

We know,

→ λ = c/ν

Or, ν = c/λ

Or, ν = (3 * 10⁸)/ (1 * 10^–10)

Or, ν = 3 * 10¹⁸ s^–1 or Hz.

So, the frequency of wavelength is 3 * 10¹⁸ s^–1 or Hz.

→ Energy, E = hν

Or, E = 4.14 * 10^–15 * 3 * 10¹⁸

Or, E = 12.4 * 10³ eV = 12.4 keV.

→ Momentum, p = E/c

Or, p = (12.4 * 10³ * 1.6 * 10^–19)/ (3 * 10⁸)

Or, p = 6.62 * 10^–24 kg m/s.

2. Iron emits K_α X-ray of energy 6.4 keV and calcium emits K_α X-ray of energy 3.69 keV. Calculate the times taken by an iron K_α photon and a calcium K_α photon to cross through a distance of 3 km.

Sol:

Given: distance, S = 3 km = 30000 m.

Speed of both the photon are same and equal to speed of light, c = 3 * 10⁸ m/s.

So, the times taken by both = S/c = 3000/ (3 * 10⁸) = 10 µs.

3. Find the cut-off wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV.

Sol:

Given: potential, V = 30 kV = 30000 V.

We know,

→ λ_min = hc/eV

Or, λ_min = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (30000 eV)

Or, λ_min = 41.4 * 10^–12 m = 41.4 pm.

4. What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

Sol:

Given: λ_min = 0.10 nm = 0.1 * 10^–9 m.

We know,

→ λ_min = hc/eV

Or, V = hc/ (eλ_min)

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 0.1 * 10^–9)

Or, V = 12.4 * 10³ V = 12.4 kV.

→ Maximum energy, E = eV

Or, E = 1.6 * 10^–19 * 12.4 * 10³

Or, E = 1.98 * 10^–15 J.

5. The X-ray coming from a Coolidge tube has a cut-off wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.

Sol:

Given: λ = 80 pm = 0.8 * 10^–10 m.

We know,

→ λ = hc/E

Or, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.8 * 10^–10)

Or, E = 15.5 * 10³ eV = 15.5 keV.

So, the kinetic energy of the electrons is 15.5 keV.

6. If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease?

Sol:

We know,

Wave length, λ = hc/eV

Now, potential, V is increased by 1%.

New wave length, λ’ = hc/1.01eV = λ/1.01

→ Δλ = λ – λ’ = (0.01/1.01) λ

So, % change of wave length = (Δλ/λ)*100

Or, % change of wave length = 1/1.01

Or, % change of wave length = 0.99 = 1 %.

7. The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

Sol:

Given: distance, d = 1.5 m; wavelength, λ = 30 pm = 0.3 * 10^–10 m. 

We know, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.3 * 10^–10)

Or, E = 41.4 * 10³ eV

Or, potential, V = E/e = 41.4 * 10³ V

Now, Electric field = V/d

Or, Electric field = 41.4 * 10³/1.5

Or, Electric field = 27.6 * 10³ V/m

Or, Electric field = 27.6 kV/m.

8. The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Sol:
Given: λ’ = λ – 26 * 10^–12; V’ = 1.5 V.

Original wavelength, λ = hc/eV

New wavelength, λ’ = hc/eV’

→ λV = λ’V’

Or, λV = (λ – 26 * 10^–12) * 1.5 V

Or, λ = (1.5 * 26 * 10^–12)/0.5

Or, λ = 78 * 10^–12 m.

So, original voltage, V = hc/eλ

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 78 * 10^–12)

Or, V = 15.93 * 10³ V = 15.93 kV.

9. The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?

Sol:

Given: voltage, V = 32 kV = 32000 V.

We know, λ = hc/eV

Or, λ = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (320000 eV)

Or, λ = 38.8 * 10^–12 m = 38.8 pm.

10. When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7 * 10¹⁸ Hz. Calculate the value of Planck constant from these data.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V; frequency, ν = 9.7 * 10¹⁸ Hz.

We know, E = hν

Or, eV = hν

Or, h = eV/ν

Or, h = 40 * 10³/9.7 * 10¹⁸ eV-s

Or, h = 4.12 * 10^–15 eV-s.

11. An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V.

Energy, E = eV = 40 * 10³ eV.

→ Energy utilized = (70/100) * 40 * 10³ eV.

Or, Energy utilized = 28 * 10³ eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (28 * 10³)

Or, λ = 44.35 * 10^–12 = 44.35 pm.

For 2^nd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (40 – 28) * 10³ eV.

Or, E = 84 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (84 * 10²)

Or, λ = 148 * 10^–12 = 148 pm.

For 3^rd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (12 – 8.4) * 10³ eV.

Or, E = 25.2 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (25.2 * 10²)

Or, λ = 493 * 10^–12 = 493 pm.

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