What is the ratio of the speed of an electron in the first orbit of an \(\mathrm{H}\text-\)atom to the speed of light?

The total energy of an electron in the first excited state of a hydrogen atom is about \(-3.4\) eV. Its kinetic energy in this state will be:
1. \(-6.8~\text{eV}\)
2. \(3.4~\text{eV}\)
3. \(6.8~\text{eV}\)
4. \(-3.4~\text{eV}\)

In the \(n^{th}\) orbit, the energy of an electron is \(E_{n}=-\frac{13.6}{n^2} ~\text{eV}\) for the hydrogen atom. What will be the energy required to take the electron from the first orbit to the second orbit?
1. \(10.2~\text{eV}\)
2. \(12.1~\text{eV}\)
3. \(13.6~\text{eV}\)
4. \(3.4~\text{eV}\)

If an electron in a hydrogen atom jumps from the \(3\)rd orbit to the \(2\)nd orbit, it emits a photon of wavelength \(\lambda\). What will be the corresponding wavelength of the photon when it jumps from the \(4^{th}\) orbit to the \(3\)rd orbit?

Considering the \(3^{rd}\) orbit of \(\mathrm{He}^{+}\) (Helium ion), using the non-relativistic approach, the speed of the electron in this orbit will be: (Given: \(Z=2, K = 9\times 10^{9}\), and Planck's constant, \(h= 6.6\times10^{-34}\) J-s)
1. \(2.92\times 10^{8}\) m/s
2. \(1.46\times 10^{6}\) m/s
3. \(0.73\times 10^{8}\) m/s
4. \(3.0\times 10^{8}\) m/s