An electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of a hydrogen atom. The wave number of the emitted radiations ($R =$ Rydberg's constant) will be
- A$\frac{16}{3R}$
- B$\frac{2R}{16}$
- C$\frac{3R}{16}$
- D$\frac{4R}{16}$
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View SolutionIf the energy of a hydrogen atom in the $n^{th}$ orbit is ${E_n}$,then the energy in the $n^{th}$ orbit of a singly ionized helium atom will be:
MediumAIPMT 2001
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