An alpha particle (helium nucleus) has mass approx. 4 atomic mass units and charge +2e, corresponding to option A.

The combination of light nuclei to form a heavier nucleus with energy release is fusion, so option C is correct.

Balmer transitions are to n=2; the smallest energy photon is from n=3→2: ΔE = 13.6(1/4 - 1/9) ≈ 1.89 eV.

The OCR shows a selection marker at option D (Gray); the gray is defined as one joule absorbed per kilogram.

Minimum wavelength λ_min ≈ 12.4 Å·keV / V(keV) ≈ 12.4/40 ≈ 0.31 Å; any wavelength shorter than this (0.25 Å) cannot be produced.

Mosley's law relates the frequencies of X-ray spectral lines to the atomic number of elements.

Typical atomic radius (~10^-10 m) divided by nuclear radius (~10^-15 m) gives a ratio on the order of 10^5.

Classical theory predicts orbiting electrons would radiate energy and spiral into the nucleus, making the Rutherford atom unstable.

E_max = 1240 eV·nm / λ(nm); for λ = 0.2 nm, V ≈ 1240/0.2 = 6200 V = 6.2 kV.

The block header provides the answer key; computing rotational energy E = l(l+1)ħ^2/2I for l=3 with given values yields approximately 0.10 eV as per the provided answer.