Question

Calculate the wavelength of hydrogen like atom $Fe\left( {26} \right)$ in Lyman series$\left( {{n_2} = 2,{n_1} = 1} \right)$. The value of$R = 1.097 \times {10^{ - 7}}m$.

Answer 1

Hint: We Understand that the Lyman arrangement is a hydrogen unearthly arrangement of changes and coming about bright discharge lines of the hydrogen iota as an electron goes from \[n \geqslant 2\]to $n = 1$ (where n is the key quantum number), the most reduced energy level of the electron.
The variant of the Rydberg recipe that produced the Lyman arrangement was,
$\dfrac{1}{\lambda } = {R_H}\left( {1 - \dfrac{1}{{{n^2}}}} \right)$

Complete step by step answer:
The outflow range of nuclear hydrogen has been partitioned into various unearthly arrangements, with frequencies given by the Rydberg technique. These watched phantom lines are because of the electron making advances between two energy levels in a molecule. The arrangement of the arrangement by the Rydberg recipe was significant in the advancement of quantum mechanics. The ghostly arrangements are significant in cosmic spectroscopy for identifying the presence of hydrogen and figuring red movements.
It is given that${n_1} = 1$, and${n_2} = 2$.
The Rydberg constant is$1.091 \times {10^{ - 7}}m$.
For most limited frequency in Lyman arrangement, the energy contrast in two states demonstrating change should be greatest.
Thus,
$\Rightarrow$ $\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
$\Rightarrow$ $\lambda = \dfrac{4}{3} \times \dfrac{1}{{{R_H}}}$
We can substitute the given value we get,
$ \Rightarrow \lambda = \dfrac{4}{{3 \times 1.091 \times {{10}^{ - 7}}m}}$
$\lambda = 1215.7{A^ \circ }$

Note:
We have to remember that a hydrogen molecule consists of an electron circling its core. The electromagnetic power between the electron and the atomic proton prompts a bunch of quantum states for the electron, each with its own energy. These states were envisioned by the Bohr model of the hydrogen atom as being particular circles around the core. Every energy state, or circle, is assigned by a number n. The Bohr model was later supplanted by quantum mechanics in which the electron possesses a nuclear orbital as opposed to a circle, yet the permitted energy levels of the hydrogen particle continued as before as in the prior hypothesis.
Ghastly emanation happens when an electron advances, or bounces, from a higher energy state to a lower energy state. To recognize the two expressions, the lower energy state is normally assigned as n′, and the higher energy state is assigned as n. The energy of a produced photon relates to the energy contrast between the two states. Since the energy of each state is fixed, the energy contrast between them is fixed, and the progress will consistently create a photon with a similar energy.