Consider a hydrogen atom with v_k, r_k, and K_k denoting the velocity, orbital radius and kinetic energy of the electron in k^th orbit, respectively. The electron undergoes a transition from the n^th orbit, emitting radiation corresponding to the Lyman series. Considering h to be the Planck’s constant and ε₀ the permittivity of free space, the correct statement(s) is/are:

The correct statements are:
1. Magnitude of change in kinetic energy can be expressed as
$\frac{h}{4\pi }|\frac{n{v}_n}{r_n}-\frac{v_1}{r_1}|$
2. Frequency of radiation emitted can be expressed as
$\frac{e^2}{8\pi {\epsilon }_{0}h}(\frac{1}{r_1}-\frac{1}{r_{\mathrm{n}}})$

Step-by-Step Derivation and Analysis:

Step 1: Expression for Kinetic Energy using Bohr's Quantization Condition
According to Bohr's postulate of angular momentum quantization for the k^th orbit of a hydrogen atom:
$m{v}_k{r}_k=\frac{kh}{2\pi }$
where m is the mass of the electron, v_k is its orbital speed, and r_k is the orbital radius.

The kinetic energy K_k in the k^th orbit is defined as:
$K_k=\frac{1}{2}m{v}_k^2$

We can rewrite the kinetic energy by grouping terms:
$K_k=\frac{1}{2}\left(m{v}_k{r}_k\right)\frac{v_k}{r_k}$

Substituting the angular momentum m v_k r_k = k h / (2π) into the kinetic energy equation gives:
$K_k=\frac{1}{2}\left(\frac{kh}{2\pi }\right)\frac{v_k}{r_k}=\frac{h}{4\pi }\frac{k{v}_k}{r_k}$

For a transition from the n^th orbit to the 1^st orbit (Lyman series transition), the change in kinetic energy is:
$\left|\Delta K\right|=|K_n-K_1|=\frac{h}{4\pi }|\frac{n{v}_n}{r_n}-\frac{v_1}{r_1}|$

Step 2: Frequency of the Emitted Radiation
The electrostatic force provides the necessary centripetal acceleration for the circular orbit:
$\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r_k^2}=\frac{m{v}_k^2}{r_k}$

Multiplying both sides by r_k / 2 gives the expression for kinetic energy:
$K_k=\frac{1}{2}m{v}_k^2=\frac{e^2}{8\pi {\epsilon }_0{r}_k}$

For a bound electron in a hydrogen atom, the total mechanical energy E_k is negative and equal in magnitude to the kinetic energy:
$E_k=-K_k=-\frac{e^2}{8\pi {\epsilon }_0{r}_k}$

During the transition from the n^th orbit to the 1^st orbit, the energy of the emitted photon (hν) is equal to the difference in energy levels:
$h\nu =E_n-E_1=(-\frac{e^2}{8\pi {\epsilon }_0{r}_n})-(-\frac{e^2}{8\pi {\epsilon }_0{r}_1})$
$h\nu =\frac{e^2}{8\pi {\epsilon }_0}(\frac{1}{r_1}-\frac{1}{r_n})$

Dividing by Planck's constant h, the frequency ν of the emitted radiation is:
$\nu =\frac{e^2}{8\pi {\epsilon }_{0}h}(\frac{1}{r_1}-\frac{1}{r_{\mathrm{n}}})$