Question

In Bohr's atomic model, the energy of the electron is 'E' in the second orbit of hydrogen atom. So the energy of the electron in the third orbit of helium atom \((Z = 2)\) will be

• A. \(\frac{16\,E}{3}\)
• B. \(\frac{16\,E}{9}\)
• C. \(\frac{4\,E}{3}\)
• D. \(\frac{4\,E}{9}\)

Hint:

Count the possible transitions from n = 3 to lower levels.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The second excited state is \(n = 3\) (ground state \(n = 1\), first excited \(n = 2\)).

Step 2: Transitions:
The electron can go \(3\to2\), \(3\to1\) and \(2\to1\). That is 3 different lines.
Using the formula \(\frac{n(n - 1)}{2} = \frac{3\times2}{2} = 3\).

Final Answer:
The maximum number of spectral lines is 3, option (B). \[ \boxed{3} \]