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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 2

Problem

Compute the potential difference through which an electron must be accelerated such that the short-wave limit of the continuous X-ray spectrum is exactly 0.1 nm0.1 \, \mathrm{nm}.

Data

  • Wavelength (λ\lambda): 0.1×10−9 m0.1 \times 10^{-9} \, \mathrm{m}
  • 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:

  • Potential difference (ΔV\Delta V).

Prerequisite Concepts

  1. Energy of a Photon (Planck’s Hypothesis):
    The energy of a photon is given by:
E=hcλ E = \frac{hc}{\lambda}

Where:

  • hh: Planck’s constant
  • cc: Speed of light
  • λ\lambda: Wavelength of the photon
  1. Electron Energy from Acceleration:
    The energy gained by an electron when accelerated through a potential difference is:
E=qΔV E = q \Delta V

Where:

  • qq: Charge of the electron
  • ΔV\Delta V: Potential difference
  1. Relationship Between Photon Energy and Potential Difference:
    Equating the two expressions for energy:
qΔV=hcλ q \Delta V = \frac{hc}{\lambda}

Solving for ΔV\Delta V:

ΔV=hcqλ \Delta V = \frac{hc}{q \lambda}

Solution

Step 1: Write the formula for potential difference

ΔV=hcqλ\Delta V = \frac{hc}{q \lambda}

Step 2: Substitute the known values

h=6.626×10−34 Js,c=3×108 m/s,q=1.602×10−19 C,λ=0.1×10−9 mh = 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 \lambda = 0.1 \times 10^{-9} \, \mathrm{m} ΔV=6.626×10−34⋅3×1081.602×10−19⋅0.1×10−9\Delta V = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^{8}}{1.602 \times 10^{-19} \cdot 0.1 \times 10^{-9}}

Step 3: 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⋅0.1×10−9=1.602×10−291.602 \times 10^{-19} \cdot 0.1 \times 10^{-9} = 1.602 \times 10^{-29}

Step 4: Calculate ΔV\Delta V

ΔV=1.9878×10−251.602×10−29\Delta V = \frac{1.9878 \times 10^{-25}}{1.602 \times 10^{-29}} ΔV=12412.5 V\Delta V = 12412.5 \, \mathrm{V}

Round to significant figures:

ΔV=12400 V\Delta V = 12400 \, \mathrm{V}

Answer

The potential difference required is:

ΔV=12400 V\Delta V = 12400 \, \mathrm{V}