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Notely Physics 12
Class-12 Physics

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9 - N u m e r i c a l   1 1

Problem

Electrons in an X-ray tube are accelerated through a potential difference of 3000 V3000 \, \mathrm{V}. If these electrons are slowed down in a target, what will be the minimum wavelength of X-rays produced?

Data

  • Potential difference (ΔV\Delta V): 3000 V3000 \, \mathrm{V}
  • Planck’s constant (hh): 6.626×10−34 Js6.626 \times 10^{-34} \, \mathrm{Js}
  • Charge on electron (qq): 1.602×10−19 C1.602 \times 10^{-19} \, \mathrm{C}
  • Speed of light (cc): 3×108 m/s3 \times 10^{8} \, \mathrm{m/s}

To Find:

  • Minimum wavelength (λ\lambda) of X-rays.

Prerequisite Concepts

  1. Planck’s Hypothesis:
    The energy of a photon is given by:
E=hcλ E = \frac{h c}{\lambda}

Where:

  • EE: Energy of the photon
  • hh: Planck’s constant
  • cc: Speed of light
  • λ\lambda: Wavelength
  1. Energy from Accelerated Electrons:
    The energy gained by electrons due to a potential difference is:
E=qΔV E = q \Delta V

Where:

  • qq: Charge on the electron
  • ΔV\Delta V: Potential difference
  1. Relationship Between Energy and Wavelength:
    Equating the two expressions for energy:
qΔV=hcλ q \Delta V = \frac{h c}{\lambda}

Solving for λ\lambda:

λ=hcqΔV \lambda = \frac{h c}{q \Delta V}

Solution

Step 1: Substitute the known values into the formula

λ=hcqΔV\lambda = \frac{h c}{q \Delta V}

Substitute:

h=6.626×10−34 Js,c=3×108 m/s,q=1.602×10−19 C,ΔV=3000 Vh = 6.626 \times 10^{-34} \, \mathrm{Js}, \quad c = 3 \times 10^{8} \, \mathrm{m/s}, \quad q = 1.602 \times 10^{-19} \, \mathrm{C}, \quad \Delta V = 3000 \, \mathrm{V} λ=6.626×10−34⋅3×1081.602×10−19⋅3000\lambda = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{1.602 \times 10^{-19} \cdot 3000}

Step 2: Simplify the numerator and denominator

Numerator:

6.626×10−34⋅3×108=1.9878×10−256.626 \times 10^{-34} \cdot 3 \times 10^{8} = 1.9878 \times 10^{-25}

Denominator:

1.602×10−19⋅3000=4.806×10−161.602 \times 10^{-19} \cdot 3000 = 4.806 \times 10^{-16}

Step 3: Calculate λ\lambda

λ=1.9878×10−254.806×10−16\lambda = \frac{1.9878 \times 10^{-25}}{4.806 \times 10^{-16}} λ=4.14×10−10 m\lambda = 4.14 \times 10^{-10} \, \mathrm{m}

Convert to angstroms (1 A˚=10−10 m1 \, \mathrm{Å} = 10^{-10} \, \mathrm{m}):

λ=4.14 A˚\lambda = 4.14 \, \mathrm{Å}

Answer

The minimum wavelength of X-rays produced is:

λ=4.14×10−10 m or 4.14 A˚\lambda = 4.14 \times 10^{-10} \, \mathrm{m} \, \text{or} \, 4.14 \, \mathrm{Å}