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Atoms and Nuclei - X-Rays

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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X-rays are electromagnetic radiations of very short wavelength, ranging from 0.1 A˚0.1 \text{ \AA} to 100 A˚100 \text{ \AA}. They are produced when high-speed electrons are suddenly decelerated by a metal target.

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The production of X-rays typically occurs in a Coolidge Tube, where electrons are emitted from a heated filament (cathode) and accelerated towards a high-atomic-number target (anode) using a high potential difference VV.

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Continuous X-ray Spectrum: This is produced due to the deceleration of electrons as they penetrate the target material (Bremsstrahlung). It has a definite minimum wavelength λmin\lambda_{min} (cutoff wavelength) which depends only on the accelerating voltage VV.

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Characteristic X-ray Spectrum: This consists of sharp peaks at specific wavelengths superimposed on the continuous spectrum. They occur when high-speed electrons knock out inner-shell electrons (like KK or LL shells) of the target atoms, and electrons from higher shells drop down to fill the vacancies.

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Moseley's Law: The square root of the frequency ν\nu of a characteristic X-ray line is directly proportional to the atomic number ZZ of the target element. This is expressed as ν=a(Z−b)\sqrt{\nu} = a(Z - b). This law established that the atomic number, not the atomic weight, is the fundamental property of an element.

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Properties of X-rays: They travel in straight lines at the speed of light cc, are not deflected by electric or magnetic fields (indicating they are neutral), ionize gases, and exhibit phenomena like diffraction and interference.

📐Formulae

E=eV=hνmax=hcλminE = eV = h\nu_{max} = \frac{hc}{\lambda_{min}}

λmin=hceV≈12420V A˚ (where V is in Volts)\lambda_{min} = \frac{hc}{eV} \approx \frac{12420}{V} \text{ \AA} \text{ (where } V \text{ is in Volts)}

ν=a(Z−b)\sqrt{\nu} = a(Z - b) (Moseley's Law)

Ephoton=Ei−EfE_{photon} = E_i - E_f (Energy of characteristic X-ray photon)

2dsin⁡θ=nλ2d \sin \theta = n\lambda (Bragg's Law for X-ray diffraction)

💡Examples

Problem 1:

An X-ray tube operates at an accelerating potential of 50 kV50 \text{ kV}. Calculate the minimum wavelength of the X-rays produced. (Take h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}, and e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C})

Solution:

Given V=50 kV=50×103 VV = 50 \text{ kV} = 50 \times 10^3 \text{ V}. Using the formula: λmin=hceV\lambda_{min} = \frac{hc}{eV} λmin=6.63×10−34×3×1081.6×10−19×50×103\lambda_{min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 50 \times 10^3} λmin=19.89×10−2680×10−16=0.2486×10−10 m\lambda_{min} = \frac{19.89 \times 10^{-26}}{80 \times 10^{-16}} = 0.2486 \times 10^{-10} \text{ m} $$ \lambda_{min} \approx 0.249 \text{ \AA}

Explanation:

The minimum wavelength (cutoff wavelength) corresponds to the case where the entire kinetic energy of the electron is converted into a single X-ray photon upon collision with the target.

Problem 2:

The KαK_\alpha X-ray frequency for a certain element is 1.88×1018 Hz1.88 \times 10^{18} \text{ Hz}. If for the KK series, the screening constant b=1b = 1, find the atomic number ZZ of the element. (Assume a≈4.97×107 Hz1/2a \approx 4.97 \times 10^{7} \text{ Hz}^{1/2} for the KαK_\alpha line transition)

Solution:

Using Moseley's Law: ν=a(Z−b)\sqrt{\nu} = a(Z - b). Given ν=1.88×1018 Hz\nu = 1.88 \times 10^{18} \text{ Hz}, b=1b = 1, and a=4.97×107a = 4.97 \times 10^7. 1.88×1018=4.97×107(Z−1)\sqrt{1.88 \times 10^{18}} = 4.97 \times 10^7 (Z - 1) 1.371×109=4.97×107(Z−1)1.371 \times 10^9 = 4.97 \times 10^7 (Z - 1) Z−1=1.371×1094.97×107≈27.58Z - 1 = \frac{1.371 \times 10^9}{4.97 \times 10^7} \approx 27.58 Z≈28.58Z \approx 28.58 (Rounding to the nearest integer, the element is Nickel with Z=28Z=28 or Copper with Z=29Z=29 depending on the specific constant aa used).

Explanation:

Moseley's Law relates the frequency of characteristic X-rays to the atomic number. For KαK_\alpha lines, bb is approximately 11 because one 1s1s electron remains to screen the nucleus.

X-Rays - Revision Notes & Key Formulas | ICSE Class 12 Physics