This question is about electrons and photons - OCR - A-Level Physics A - Question 8 - 2012 - Paper 1

The energy gained by an electron accelerated through a potential difference (p.d.) is given by

$E = eV$

where:

• $E$ is the energy in joules,
• $e$ is the charge of the electron ($1.6 imes 10^{-19}$ C),
• $V$ is the potential difference (5000 V).

Substituting the values:

$E = (1.6 imes 10^{-19} ext{ C})(5000 ext{ V}) = 8.0 imes 10^{-16} ext{ J}$

The kinetic energy gained by the electron as it accelerates can be equated to its kinetic energy:

E_k = rac{1}{2} mv^2

where:

• $E_k$ is the kinetic energy ($8.0 imes 10^{-16} ext{ J}$),
• $m$ is the mass of the electron ($9.11 imes 10^{-31} ext{ kg}$), and
• $v$ is the speed.

Rearranging and substituting:

ho{2E_k}{m}$$ $$v = ho{2(8.0 imes 10^{-16} ext{ J})}{9.11 imes 10^{-31} ext{ kg}} \\ v \\approx 4.0 imes 10^{7} ext{ ms}^{-1}$$

The de Broglie wavelength of an electron represents the wavelength associated with a moving electron, effectively describing its wave-particle duality. This concept states that every matter particle, including electrons, can be associated with a wavelength, which can be calculated using the formula:

$\\lambda = \frac{h}{p}$

where:

• $\lambda$ is the de Broglie wavelength,
• $h$ is Planck's constant ($6.626 imes 10^{-34} ext{ Js}$), and
• $p$ is the momentum of the particle.

To calculate the de Broglie wavelength, first, we need the momentum of the electron, given by:

$p = mv$

With $m = 9.11 imes 10^{-31} ext{ kg}$ and $v \\approx 4 x 10^7 ext{ ms}^{-1}$:

$p = (9.11 imes 10^{-31} ext{ kg})(4 imes 10^7 ext{ ms}^{-1}) \\approx 3.644 imes 10^{-23} ext{ kg m/s}$

Now substituting into the de Broglie equation:

$\\lambda = \frac{h}{p} = \frac{6.626 imes 10^{-34} ext{ Js}}{3.644 imes 10^{-23} ext{ kg m/s}} \\approx 1.82 imes 10^{-11} ext{ m}$