One of the fundamental laws of physics is that matter is most stable with the lowest…
Chemistry · JEE Advanced · NTA Exams — Atomic Structure
One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n = 1 orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher n value, in which case the atoms is in an excited with a higher energy.
The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state.
Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively.
When an electron returns from n2 to n1 state, the number of lines in the spectrum will equal to
\(\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}\)
If the electron comes back from energy level having energy E2 to energy level having energy, E1, then the difference may be expressed in terms of energy of photon as :
\(\mathrm{E}_{2}-\mathrm{E}_{1}=\Delta \mathrm{E}, \Delta \mathrm{E} \Rightarrow \frac{\mathrm{hc}}{\lambda}\)
Since, h and c are constants, ∆E corresponds to definite energy ; thus, each transition from one energy level to another with produce a radiation of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom.
Wave number of a spectral line is given by the formula
\(\overline{\mathrm{v}}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\)
where R is a Rydberg’s constant (R = 1.1 × 107 m–1)
What transition in the hydrogen spectrum would have the same wavelength as Balmer transition, n = 4 to n = 2 in the He+ spectrum ?
- n = 3 to n = 1
- n = 3 to n = 2
- n = 4 to n = 1
- n = 2 to n = 1
Answer
(D) n = 2 to n = 1
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