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Dual Nature of Radiation question

2008 · Shift 1 · Q49
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Dual Nature of Radiation question

2008 · Shift 1 · Q49

JEE AdvancedPhysicsDual Nature of RadiationMCQ+3 / −1
Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube?
  1. A
    Wavelength of characteristic X-rays decreases when the atomic number of the target increases.
  2. B
    Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.
  3. C
    Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube.
  4. D
    Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube.
View written solutionFree

Correct answer: B

To identify the incorrect statement regarding X-rays generated from an X-ray tube, we need to analyze the physics behind the production of both continuous and characteristic X-rays.

Step 1: Analyze Continuous X-rays (Bremsstrahlung)

Continuous X-rays are produced when high-energy electrons are rapidly decelerated upon striking the metal target. The maximum energy (EmaxE_{max}Emax​) of an emitted X-ray photon corresponds to the case where an electron loses all its kinetic energy (KKK) in a single collision.

The kinetic energy of an electron accelerated through a potential difference VVV is given by: K=eVK = eVK=eV where eee is the elementary charge.

The maximum energy of the emitted photon is thus: Emax=eVE_{max} = eVEmax​=eV The relationship between the energy of a photon (EEE) and its wavelength (λ\lambdaλ) is: E=hcλE = \frac{hc}{\lambda}E=λhc​ where hhh is Planck's constant and ccc is the speed of light.

The shortest possible wavelength, known as the cut-off wavelength (λmin\lambda_{min}λmin​), corresponds to the maximum photon energy: λmin=hcEmax=hceV\lambda_{min} = \frac{hc}{E_{max}} = \frac{hc}{eV}λmin​=Emax​hc​=eVhc​ From this equation, we can see that the cut-off wavelength λmin\lambda_{min}λmin​ depends only on the accelerating voltage VVV and not on the material of the target (i.e., not on its atomic number ZZZ).

Step 2: Analyze Characteristic X-rays

Characteristic X-rays are produced when an incident high-energy electron knocks out an electron from an inner shell (e.g., K-shell) of a target atom. An electron from a higher energy shell then transitions to fill the vacancy, emitting a photon with an energy equal to the difference between the two energy levels. These energy levels are specific to the atoms of the target material.

Moseley's Law relates the frequency (fff) of a characteristic X-ray line (like Kα_{\alpha}α​) to the atomic number (ZZZ) of the target element: f=a(Z−b)\sqrt{f} = a(Z-b)f​=a(Z−b) where aaa and bbb are constants. Since E=hfE = hfE=hf and E=hc/λE = hc/\lambdaE=hc/λ, we can write: E∝(Z−b)2E \propto (Z-b)^2E∝(Z−b)2 1λ∝(Z−b)2\frac{1}{\lambda} \propto (Z-b)^2λ1​∝(Z−b)2 This implies that as the atomic number ZZZ of the target increases, the energy of the characteristic X-rays increases, and consequently, their wavelength λ\lambdaλ decreases.

Step 3: Analyze X-ray Intensity

The intensity of X-rays refers to the energy per unit area per unit time, which is related to the number of photons produced per second. The number of photons is proportional to the number of electrons hitting the target per second, which is the tube current (III). The electrical power supplied to the X-ray tube is P=VIP = VIP=VI. Increasing either the voltage VVV or the current III increases the power. An increase in current leads to more electrons hitting the target, thus producing more photons (both continuous and characteristic), which increases the intensity. An increase in voltage increases the energy of the electrons, which also increases the efficiency of X-ray production and hence the intensity.

Step 4: Evaluate each statement

A: Wavelength of characteristic X-rays decreases when the atomic number of the target increases. As shown by Moseley's Law (Step 2), λ∝1/(Z−b)2\lambda \propto 1/(Z-b)^2λ∝1/(Z−b)2. So, as ZZZ increases, λ\lambdaλ decreases. This statement is CORRECT.

B: Cut-off wavelength of the continuous X-rays depends on the atomic number of the target. As shown in Step 1, the cut-off wavelength is given by λmin=hc/eV\lambda_{min} = hc/eVλmin​=hc/eV. It depends on the accelerating voltage VVV, but not on the atomic number ZZZ of the target. This statement is WRONG.

C: Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube. As discussed in Step 3, the intensity is proportional to the tube current and also depends on the accelerating voltage. Both of these factors contribute to the electrical power (P=VIP=VIP=VI). Therefore, intensity depends on the electrical power. This statement is CORRECT.

D: Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube. As shown in Step 1, λmin=hc/Emax\lambda_{min} = hc/E_{max}λmin​=hc/Emax​, where EmaxE_{max}Emax​ is the maximum kinetic energy of the electrons. Thus, the cut-off wavelength is inversely proportional to the energy of the electrons. This statement is CORRECT.

Conclusion

The question asks for the WRONG statement. Based on our analysis, statement B is the incorrect one.

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