# HC Verma Solutions Class 12 Chapter 22 X-Rays

HC Verma Solutions Class 12 Chapter 22 X-Rays offers detailed answers to all the exercises given in the chapter. Students can, therefore, use the solution to not only know the correct answers but they will get a clear understanding of the right method to solve these questions. The solutions will further help them develop better problem-solving skills and ultimately boost their preparation level. The solutions are centred around physics topics such as frequency, momentum and the energy of X-rays, etc. StudentsÂ will also get to learn topics of X-Rays in problems such as:

• Finding the electric field when wavelength and diameter are given.
• Finding energy utilised by X-Rays when the wavelength is given.
• Questions based on energy needed to knock out an electron from their shells and energy needed to ionize an atom, etc.
• We will be seeing questions in which we have to find the maximum potential difference without emitting an electron, the power emitted in X-ray, heat produced, stopping potential and momentum of the atom.

The HC Verma solutions which have been thoughtfully prepared will also help students to be familiar with the concepts and practice the questions in a seamless manner.

## Key Topics Related To X-rays

Some of the key topics that students should focus on are;

• Production of X-rays
• Continuous and Characteristic X-rays
• Soft and Hard X-Rays
• Moseley’s Law
• Bragg’s Law
• Properties and Uses of X-rays

## Class 12 Important Questions In Chapter 22

1. State whether X-rays can be used or not for photoelectric effect? Is it possible for X-rays to be polarized?

2. Electrons usually strike the target and stop inside a Coolidge tube. Explain whether the target gets negatively charged as time passes in the process? Â

3. It is often said that X-ray is injurious to health. However, exposure to visible light is not harmful. Explain why this happens when both of them are electromagnetic waves? Â

4. What is the nature of the X-ray beam coming from an X-ray tube?

(a) It is monochromatic (b) the wavelengths are smaller than a certain maximum wavelength (c) the wavelengths are greater (d) All the wavelengths lie between the minimum and maximum wavelength

5. What happens to the emitted radiation when the potential difference applied to an X-ray tube is increased?

(a) the minimum wavelength decreases (b) the minimum wavelength increases (c) the intensity increases (d) there is no change in the level of intensity

## HC Verma Solutions Vol 2 X-Rays Chapter 22

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

Solution:

Wavelength of X-rays = 0.10 nm.

We know, E = hc/Î»

Where,

Velocity of light = c = 3 x 108 m/s

Planks constant = h = 6.63 x 10-34 J-s

Wavelength of X-rays = Î» = 0.10 x 10-9 m (given)

E = [6.63 x 10-34 x 3 x 108 ]/[ 0.10 x 10-9] = 1.98 x 10-15 Joules/photon

Also, the frequency can be calculated using below formula:

Î³ = c/ Î» = [3 x 108]/[ 0.10 x 10-9] = 3 x 1018 Hertz

Find momentum:

p = h/ Î» = [6.63 x 10-34]/[ 0.10 x 10-9] = 6.626 x 10-24 kg ms-1.

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

Solution:

Let “t” be the time taken each of those rays to travel a distance of 3 Km.

We know, time = distance/speed

=> t = [3×103]/[3×108] = 10-5 s = 10 Î¼s

Both the KÎ± photon and X-ray photon will take the same i.e. 10Î¼s

Question 3: Find the cutoff wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV.

Solution:

Cutoff wavelength = Î» = hc/eV = [1242 eV-nm]/[ex30x103] = 414 x 10-4 nm

Question 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?

Solution:

We know, Î» = hc/eV

Or V = hc/eÎ» = [6.63 x 10-34 x 3 x108]/[1.6×10-19x10-10] = 12.4 KV

Now,

Maximum energy of photon of wavelength Î» = 0.10 nm is

E = hc/Î» = [6.63 x 10-34 x 3 x108]/[0.10×10-9] = 1.98 x 10-15 Joules/photon

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

Solution:

We know, E = hc/Î» = [1242 eV-nm]/[80×10-3] = 15.525 x 103 eV = 15.5 KeV(approx..)

Question 6: If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cut-off wavelength decrease?

Solution:

If the operating voltage is increased by 1%, then the new operating voltage, say v’ will be,

V’ = V + V/100 = 1.01 V

Cutoff wavelength (Î»’) in increasing the operation voltage is

Î»’ = (h/1.01) V = Î»/1.01

Therefore, difference in wavelength is Î»’ – Î»/1.01 = (0.01)/1.01Î»

Now,

Percentage change in the wavelength:

(0.01)/1.01Î» x Î» x 100 = 1/1.01 = 0.9901 = 1 % (approx)

Question 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.

Solution:

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.

We know, E = hc/ Î»

= [6.63 x 10-34 x 3 x108]/[30×10-12] = 41.4 x 103 eV

Now,

Electric field = V/d = [41.4 x 103]/1.5 = 27.6 kVm-1.

Question 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?

Solution:

When the operating voltage of the X-ray tube is increased to 1.5 times

we have, Î»’ = Î» – 26

where Î» = initial wavelength and Î»’ = new wavelength

Also, we have E = hc/Î»

or eV = hc/Î»

Where V = operating initial potential

Î» = hc/eV

Consider Vâ€™ be the new operating voltage, then

Î»V = Î»’V’ = (Î» – 26) x 1.5 V

=> 0.5Î» = 26 x 1.5

=> Î» = 78 pm

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

Solution:

E = hc/ Î»

Or Î» = hc/E

Given: Potential across the X-ray Tube, V = 32kV

=> Î» = hc/E = [1242×10-9]/[32×103] = 38.8 pm

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

Solution:

Wavelength of X-ray = Î» = hc/eV â€¦.(1)

We know, Î³ = c/ Î», using in (1), we get

h = eV/ Î³ = [40×103]/[9.7×1018] x e = 4.12 x 10-15 eVs

Question 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.

Solution:

Energy utilized by the electron = E = 70% energy into a photon

=> E = 70/100 x 40 x 103 eV = 28 x 103 eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10-9]/[28×103]

= 44.35 pm

For the 2nd wavelength (which is 70% of leftover energy)

E = 70/100 x (40-28) x 103 eV = 8.4 x 103 eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10-9]/[8.4×103]

= 148 pm

Similarly, For the 3rd wavelength:

E = 70/100 x (12-8.4) x 103 eV = 25.2 x 102 eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10-9]/[25.2×102]

= 493 pm

Question 12: The wavelength of KÎ± X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of KÎ± X-ray?

Solution:

Energy required to knock out an electron from L-shell = EL = 11.3 KeV

Wavelength of X-ray = 21.3 pm

Voltage = 11.3 kV

The energy gap between K and L shell,

EK – EL = [1242×10-9]/[21.3×10-12] = 58.309 KeV

Ek = 11.3 + 58.309 = 69.609 KeV

The accelerating voltage across the X-ray tube for the production of KÎ± X-ray is Vk = 69.609 KeV

Question 13: The KÎ² X-ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom

Solution:

Energy of KÎ² x-ray of argon, E = [1242×10-9]/[0.36×10-9] = 3450 eV

[Given: Wavelength of KÎ² X-ray of argon = 0.36 nm]

Energy needed to knock out an electron from K shell can be calculated as:

Ek = 3450 + 16 eV

Ek = 3.47 KeV (approx.)

Question 14: The KÎ± X-rays of aluminum (Z = 13) and zinc (Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseleyâ€™s law âˆšv = a(Z â€“ b) to find the wavelength of the KÎ± X-ray of iron (Z = 26).

Solution:

Let Î³1 and Î³2 are the frequencies of KÎ± X-rays of aluminum and zinc respectively

So, Î³1 = [3×108]/[887 x 10-12] = 33.82 x 1016 Hz

and Î³2 = [3×108]/[146 x 10-12] = 2.055 x 1018 Hz

Again,

âˆšÎ³ = a(Z – b)

For aluminum:

5.815 x 108 = a(13 – b) …(a)

For Zinc:

1.4331 x 109 = a(30 – b) …(a)

Dividing above equations, we get

(13-b)/(30-b) = [5.815 x 108 ]/[1.4331 x 109 ] = 0.4057

=> (13-b)/(30-b) = 0.4057

=> (13-b)0.4057 = (30-b)

=> b = 1.3949

(a)=> a = [5.815 x 108]/(13-1.3949) = 5 x 107

Now, let us find the wavelength of the KÎ± X-ray of iron:

Frequency of the iron is given as âˆšÎ³ = (5 x 107)(26 – 1.39) = 123.05 x 107

or Î³ = 5.1413 x 1014

We know, Î³ = c/Î»

or Î» = c/Î³ = [3×108]/[5.1413 x 1014] = 198 pm

Question 15: A certain element emits KÎ± X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

Solution:

Energy of KÎ± X-rays= 3.69 KeV (Given)

Using formula to find the Wavelength, or Î» = hc/eV

=> Î» = [1242×10-9]/[3690] = 0.34 x 10-9 m

[Because 3.69 KeV = 3690 eV]

Using Moseleyâ€™s law, âˆšÎ³ = a(Z – b) ….(1)

Where Î³ = c/Î»

âˆšÎ³ = âˆšc/Î» = âˆš[3×108]/âˆš[0.34×10-9] = 9.39 x 108

(1)=> 9.39 x 108 = (5 x 107)(Z – 1.39)

[Note: Used values of a and b from previous problem]

=> z = 20.17 = 20 (approx)

Atomic number (Z) = 20

Therefore, element with Atomic number 20 is calcium (Ca).

Question 16: The KÎ² X-rays from certain elements are given below. Draw a Moseley-type plot of âˆšv versus Z for KÎ² radiation.

 Element Ne P Ca Mn Zn Br Energy(KeV) 0.858 2.14 4.02 6.51 9.57 13.3

Solution:

Using energy frequency relation to find the value of v.

i.e. Frequency = Energy/h

Where h = Planck’s constant

and energy in KeV.

The required graph is:

Question 17: Use Moseleyâ€™s law with b = 1 to find the frequency of the KÎ± X-ray of La (Z = 57) if the frequency of the KÎ± X-ray of Cu (Z = 29) is known to be 1.88 Ã— 1016 Hz.

Solution:

Using Moseleyâ€™s law, âˆšÎ½ = a(Z – b) ….(1)

Where Î½ = c/Î»

Given b = 1

Now,

Given: ZLa = 57 and ZCu = 29

=> Î½La = 1.88 x 1018 ((57-1)/(29-1))2 = 7.52 x 1018

=> Î½La = 7.52 x 1018 Hz

Question 18: The KÎ± and KÎ² X-rays of molybdenum have wavelengths 0.71 angstrom and 0.63 angstrom respectively. Find the wavelength of LÎ± X-ray of molybdenum.

Solution: The KÎ± and KÎ² X-rays of molybdenum have wavelengths 0.71 angstrom and 0.63 angstrom respectively.

Let Î»1 = 0.71 angstrom and Î»2 = 0.63 angstrom

To find: Wavelength of LÎ±

Energy of KÎ± x-rays is given by: KÎ± = EK – EL

Energy of KÎ² is given by: KÎ² = EK – EM and

Energy of LÎ± is given by: LÎ± = EK – EM

Relating above equations, we have

LÎ± = KÎ± – KÎ²

= [3×108]/[0.63×10-10] – [3×108]/[0.71×10-10]

or LÎ± = 0.536 x 1018 Hz

We know, Î» = c/v = (3×108)/(0.536×1018) = 5.6 angstrom.

Question 19: The wavelengths of KÎ± and LÎ± X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of KÎ² X-ray of the material.

Solution:

The wavelengths of KÎ± and LÎ± X-rays of a material are 21.3 pm and 141 pm respectively.

Let E1, E2 and E3 are the energies of KÎ±, LÎ± and KÎ².

Using all three relations, we have

E3 = E1 + E2

=> E3 = 1242/[21.3×10-3] + 1242/[141×10-5]

= 58.309 x 103 + 8.8085 x 103

= 67.118 x 103 eV

Now, Find the wavelength of KÎ² X-ray of the material:

Using relation,

Î» = hc/E = (1242)/(67.118 x 103) = 18.5 pm.

Question 20: The energy of a silver atom with a vacancy in K shell is 25.31 keV, in L shell is 3.56 keV and in M shell is 0.530 keV higher than the energy of the atom with no vacancy. Find the frequency of KÎ±, LÎ² and LÎ± X-rays of silver.

Solution:

Let E1, E2 and E3 are the energies of K shell, L shell and M shell.

So, E1 = 25.31 KeV , E2 = 3.56 keV and E3 = 0.530 keV

Let v1, v2 and v3 be the frequencies of KÎ±, LÎ² and LÎ± x-rays respectively.

KÎ± x-ray emitted when the transition takes place between l and k shells.

=> KÎ± = E1 – E2 = h v1

or v1 = (E1 â€“ E2)/h = (25.31-3.56)/(6.63×10-34) x 1.6 x 10-19 x 103

= 5.25 x 1018 Hz

kÎ² x-ray is emitted when the transition takes place between k and m shells

=> KÎ² = E1 â€“ E3 = h v2

or v2 = (E1 â€“ E3)/h = (25.31-0.53)/(6.63×10-34) x 1.6 x 10-19 x 103

= 5.985 x 1018 Hz

LÎ± x-ray is emitted when the transition takes place between l and m shells

=> LÎ± = E2 â€“ E3 = h v3

or v3 = (E2 â€“ E3)/h = (3.56-0.53)/(6.63×10-34) x 1.6 x 10-19 x 103

= 7.32 x 1017 Hz