Placed 1 m apart, two initially neutral particles undergo a process...

When charge is transferred between two initially neutral particles, one acquires a charge $+q$ and the other $-q$. The charge must be an integer multiple of the elementary charge, so $q = ne$ where $n = 1, 2, 3, \dots$

The initial electric potential energy of the system is zero. The final electric potential energy is:

$U_f = -\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{r} = -\dfrac{n^2 e^2}{4\pi\epsilon_0 r}$

The lost electric potential energy is converted into a photon of wavelength $\lambda$. Thus, we have:

$\Delta U = 0 - U_f = \dfrac{n^2 e^2}{4\pi\epsilon_0 r}$

Equating this to the energy of the photon, $\dfrac{hc}{\lambda}$:

$\dfrac{hc}{\lambda} = \dfrac{n^2 e^2}{4\pi\epsilon_0 r} \Rightarrow \lambda = \dfrac{4\pi\epsilon_0 h c r}{n^2 e^2}$

The longest wavelength corresponds to the minimum energy, which occurs for $n = 1$. Substituting the standard values $h = 6.6 \times 10^{-34} \text{ J s}$, $c = 3 \times 10^8 \text{ m/s}$, $e = 1.6 \times 10^{-19} \text{ C}$, $r = 1 \text{ m}$, and $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$:

$\lambda_1 = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{9 \times 10^9 \times (1.6 \times 10^{-19})^2}$

$\lambda_1 = \dfrac{19.8 \times 10^{-26}}{23.04 \times 10^{-29}} \approx 860 \text{ m}$

The next smaller wavelength corresponds to $n = 2$:

$\lambda_2 = \dfrac{\lambda_1}{2^2} = \dfrac{860}{4} = 215 \text{ m}$