Question

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:

• A. \(V^2\)
• B. \({\sqrt{V}}\)
• C. \(\frac{1}{V}\)
• D. \(\frac{1}{\sqrt{V}}\)

Answer 1

The Correct Option is C

To determine the relationship between the minimum wavelength of X-rays produced and the potential difference through which an electron is accelerated, we need to understand the concept of X-ray production via the acceleration of electrons.

When electrons are accelerated through a potential difference \( V \), they gain kinetic energy equal to the electrical energy, given by the equation:

\(KE = eV\)

where:

• \(KE\) is the kinetic energy of the electron.
• \(e\) is the elementary charge (approximately \(1.6 \times 10^{-19}\) coulombs).
• \(V\) is the potential difference through which the electron is accelerated.

When these accelerated electrons strike a metal target, they decelerate rapidly, producing X-rays due to the conversion of their kinetic energy into photon energy. The energy of these X-ray photons is given by:

\(E = \frac{hc}{\lambda}\)

where:

• \(E\) is the energy of the X-ray photon.
• \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \ \text{Js}\)).
• \(c\) is the speed of light (approximately \(3 \times 10^8 \ \text{m/s}\)).
• \(\lambda\) is the wavelength of the X-rays.

Equating the two expressions for energy (since kinetic energy is transformed into photon energy), we have:

\(eV = \frac{hc}{\lambda}\)

Rearranging the formula to solve for the wavelength \(\lambda\), we obtain:

\(\lambda = \frac{hc}{eV}\)

This equation shows that the minimum wavelength \(\lambda\) is inversely proportional to the potential difference \(V\):

\(\lambda \propto \frac{1}{V}\)

Therefore, the correct answer, indicating the proportional relationship between the minimum wavelength of X-rays and the potential difference, is \(\frac{1}{V}\).