Read the extract and answer the questions that follow - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 2

To calculate the wavelength of the electrons, we can use the formula for the diffraction pattern:

$d imes \sin(\theta) = n\lambda$

Where:

• $d$ is the grating spacing,
• $\theta$ is the angle corresponding to the first order maximum,
• $n$ is the order (1 for the first maximum), and
• $\lambda$ is the wavelength of the electrons.

Given:

• Distance from the grating, $L = 0.24 \, m$,
• Central maximum distance to the first order maximum, $y = 55 \, \mu m = 55 \times 10^{-6} \, m$.

The angle for the first order maximum can be approximated using:

$\theta = \tan^{-1}\left(\frac{y}{L}\right)$

Substituting the values gives:

• $\theta \approx \tan^{-1}\left(\frac{55 \times 10^{-6}}{0.24}\right) \approx 0.000229 \, radians$.

Now, we also need the grating spacing, $d$. For 500 lines per mm, we have:

• $d = \frac{1}{500} \times 10^{-3} = 2 \times 10^{-6} \, m$.

Substituting these into the diffraction equation:

$2 \times 10^{-6} \times \sin(0.000229) = 1 \times \lambda$.

Solving for $\lambda$ gives:

$\lambda \approx 4.58 \times 10^{-7} \, m = 458 \nm$.

The wavelength of electrons cannot be determined using this diffraction grating because the wavelength of electrons is much smaller than the wavelength of light. In general, the de Broglie wavelength of electrons is significantly less than that of visible light, making the diffraction effects negligible when passing through the same grating. Thus, the spacing in the diffraction grating is inappropriate for resolving the much smaller electron wavelengths, which leads to no observable diffraction pattern for electrons.

Excited atoms emit light with discrete wavelengths due to the quantized energy levels within the atoms. When an atom absorbs energy, its electrons can jump to higher energy levels. However, these excited states are unstable, and the electrons will eventually return to lower energy levels, emitting photons in the process. The energy of these emitted photons corresponds to the difference in energy between the two levels involved. Since energy levels in atoms are quantized, the emitted light will have specific wavelengths associated with these transitions, resulting in discrete emission lines characteristic of the particular element.