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.

**Download Solutions As PDF: **HC Verma Solutions Chapter 22 PDF

## 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 10^{8} 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 10^{8} ]/[ 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 10^{8}]/[ 0.10 x 10^{-9}] = 3 x 10^{18} 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 K_{Î±} X-ray of energy 6.4 keV and calcium emits K_{Î±} X-ray of energy 3.69 keV. Calculate the times taken by an iron K_{Î±} photon and a calcium K_{Î±} 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 = [3×10^{3}]/[3×10^{8}] = 10^{-5} s = 10 Î¼s

Both the K_{Î±} 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]/[ex30x10^{3}] = 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 x10^{8}]/[1.6×10^{-19}x10^{-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 x10^{8}]/[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 10^{3} 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 x10^{8}]/[30×10^{-12}] = 41.4 x 10^{3} eV

Now,

Electric field = V/d = [41.4 x 10^{3}]/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}]/[32×10^{3}] = 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— 10^{18} 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 = [40×10^{3}]/[9.7×10^{18}] 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 10^{3} eV = 28 x 10^{3} eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10^{-9}]/[28×10^{3}]

= 44.35 pm

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

E = 70/100 x (40-28) x 10^{3} eV = 8.4 x 10^{3} eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10^{-9}]/[8.4×10^{3}]

= 148 pm

**Similarly, For the 3rd wavelength:**

E = 70/100 x (12-8.4) x 10^{3} eV = 25.2 x 10^{2} eV

Wavelength of the X-ray = Î» = hc/eV

= [1242×10^{-9}]/[25.2×10^{2}]

= 493 pm

**Question 12: **The wavelength of K_{Î±} 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 K_{Î±} X-ray?

**Solution: **

Energy required to knock out an electron from L-shell = E_{L} = 11.3 KeV

Wavelength of X-ray = 21.3 pm

Voltage = 11.3 kV

The energy gap between K and L shell,

E_{K} – E_{L} = [1242×10^{-9}]/[21.3×10^{-12}] = 58.309 KeV

E_{k} = 11.3 + 58.309 = 69.609 KeV

The accelerating voltage across the X-ray tube for the production of K_{Î±} X-ray is V_{k} = 69.609 KeV

**Question 13:** The K_{Î²} 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 K_{Î²} x-ray of argon, E = [1242×10^{-9}]/[0.36×10^{-9}] = 3450 eV

_{Î²}X-ray of argon = 0.36 nm]

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

E_{k} = 3450 + 16 eV

E_{k} = 3.47 KeV (approx.)

**Question 14:** The K_{Î±} 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 K_{Î±} X-ray of iron (Z = 26).

**Solution: **

Let Î³_{1} and Î³_{2} are the frequencies of K_{Î±} X-rays of aluminum and zinc respectively

So, Î³_{1} = [3×10^{8}]/[887 x 10^{-12}] = 33.82 x 10^{16} Hz

and Î³_{2} = [3×10^{8}]/[146 x 10^{-12}] = 2.055 x 10^{18 }Hz

Again,

âˆšÎ³ = a(Z – b)

For aluminum:

5.815 x 10^{8} = a(13 – b) …(a)

For Zinc:

1.4331 x 10^{9} = a(30 – b) …(a)

Dividing above equations, we get

(13-b)/(30-b) = [5.815 x 10^{8 }]/[1.4331 x 10^{9} ] = 0.4057

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

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

=> b = 1.3949

(a)=> a = [5.815 x 10^{8}]/(13-1.3949) = 5 x 10^{7}

Now, let us find the wavelength of the K_{Î±} X-ray of iron:

Frequency of the iron is given as âˆšÎ³ = (5 x 10^{7})(26 – 1.39) = 123.05 x 10^{7}

or Î³ = 5.1413 x 10^{14}

We know, Î³ = c/Î»

or Î» = c/Î³ = [3×10^{8}]/[5.1413 x 10^{14}] = 198 pm

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

**Solution: **

Energy of K_{Î±} 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

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

Where Î³ = c/Î»

âˆšÎ³ = âˆšc/Î» = âˆš[3×10^{8}]/âˆš[0.34×10^{-9}] = 9.39 x 10^{8}

(1)=> 9.39 x 10^{8} = (5 x 10^{7})(Z – 1.39)

=> z = 20.17 = 20 (approx)

Atomic number (Z) = 20

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

**Question 16:** The K_{Î²} X-rays from certain elements are given below. Draw a Moseley-type plot of âˆšv versus Z for K_{Î²} 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 K_{Î±} X-ray of La (Z = 57) if the frequency of the K_{Î±} X-ray of Cu (Z = 29) is known to be 1.88 Ã— 10^{16} Hz.

**Solution: **

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

Where Î½ = c/Î»

Given b = 1

Now,

Given: Z_{La} = 57 and Z_{Cu} = 29

=> Î½_{La} = 1.88 x 10^{18} ((57-1)/(29-1))^{2} = 7.52 x 10^{18}

=> Î½_{La} = 7.52 x 10^{18 }Hz

**Question 18:** The K_{Î±} and K_{Î²} X-rays of molybdenum have wavelengths 0.71 angstrom and 0.63 angstrom respectively. Find the wavelength of L_{Î±} X-ray of molybdenum.

**Solution:** The K_{Î±} and K_{Î²} 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 L_{Î±}

Energy of K_{Î±} x-rays is given by: K_{Î±} = E_{K} – E_{L}

Energy of K_{Î²} is given by: K_{Î²} = E_{K} – E_{M} and

Energy of L_{Î±} is given by: L_{Î±} = E_{K} – E_{M}

Relating above equations, we have

L_{Î±} = K_{Î±} – K_{Î²}

= [3×10^{8}]/[0.63×10^{-10}] – [3×10^{8}]/[0.71×10^{-10}]

or L_{Î±} = 0.536 x 10^{18} Hz

We know, Î» = c/v = (3×10^{8})/(0.536×10^{18}) = 5.6 angstrom.

**Question 19:** The wavelengths of K_{Î±} and L_{Î±} X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of K_{Î²} X-ray of the material.

**Solution: **

The wavelengths of K_{Î±} and L_{Î±} X-rays of a material are 21.3 pm and 141 pm respectively.

Let E_{1}, E_{2} and E_{3} are the energies of K_{Î±}, L_{Î±} and K_{Î²}.

Using all three relations, we have

E_{3} = E_{1} + E_{2}

=> E_{3} = 1242/[21.3×10^{-3}] + 1242/[141×10^{-5}]

= 58.309 x 10^{3} + 8.8085 x 10^{3}

= 67.118 x 10^{3} eV

Now, Find the wavelength of K_{Î²} X-ray of the material:

Using relation,

Î» = hc/E = (1242)/(67.118 x 10^{3}) = 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 K_{Î±}, L_{Î²} and L_{Î±} X-rays of silver.

**Solution: **

Let E_{1}, E_{2} and E_{3} are the energies of K shell, L shell and M shell.

So, E_{1} = 25.31 KeV , E_{2} = 3.56 keV and E_{3 }= 0.530 keV

Let v_{1}, v_{2} and v_{3} be the frequencies of K_{Î±}, L_{Î²} and L_{Î±} x-rays respectively.

**K _{Î±} x-ray emitted when the transition takes place between l and k shells.**

=> K_{Î±} = E_{1} – E_{2} = h v_{1}

or v_{1} = (E_{1} â€“ E_{2})/h = (25.31-3.56)/(6.63×10^{-34}) x 1.6 x 10^{-19} x 10^{3}

= 5.25 x 10^{18} Hz

**k _{Î²} x-ray is emitted when the transition takes place between k and m shells**

=> K_{Î²} = E_{1} â€“ E_{3} = h v_{2}

or v_{2} = (E_{1} â€“ E_{3})/h = (25.31-0.53)/(6.63×10^{-34}) x 1.6 x 10^{-19} x 10^{3}

= 5.985 x 10^{18} Hz

**L _{Î±} x-ray is emitted when the transition takes place between l and m shells**

=> L_{Î±} = E_{2} â€“ E_{3} = h v_{3}

or v_{3} = (E_{2} â€“ E_{3})/h = (3.56-0.53)/(6.63×10^{-34}) x 1.6 x 10^{-19} x 10^{3}

= 7.32 x 10^{17} Hz