When electrons are de-exciting to the ground state from nth orbit of hydrogen atoms, 15 spectral lin

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When electrons are de-exciting to the ground state from nth orbit of hydrogen atoms, 15 spectral lines are formed. The shortest wavelength among these will be

Accepted Answer

$\frac{n(n - 1)}{2} = 15 \Rightarrow n = 6$

Now, for shortest wavelength, required transition is 6 → 1.

$\therefore\frac{1}{\lambda} = R \times 1^{2}\left( \frac{1}{1^{2}} - \frac{1}{6^{2}} \right) = \frac{35}{36}R \Rightarrow \lambda = \frac{36}{35R}$

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