A gaseous excited hydrogen-like species with nuclear charge Z can emit radiations of six different photon energies.
The principal quantum number of the excited state is :

Options

A. `6`
B. `5`
C. `4`
D. `3`

Answer

To find the principal quantum number \( n \) of the excited state for a hydrogen-like species that can emit six different photon energies, we can use the formula for the number of spectral lines emitted during electronic transitions:

\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]

Given that the number of different photon energies emitted is 6, we can set up the equation:

\[ \frac{n(n-1)}{2} = 6 \]

### Step 1: Multiply both sides by 2 to eliminate the fraction

\[ n(n-1) = 12 \]

### Step 2: Rearrange the equation into standard quadratic form

\[ n^2 - n - 12 = 0 \]

### Step 3: Factor the quadratic equation

We need to find two numbers that multiply to \(-12\) and add to \(-1\). The numbers \(-4\) and \(3\) satisfy this:

\[ (n - 4)(n + 3) = 0 \]

### Step 4: Solve for \( n \)

Setting each factor equal to zero gives us:

1. \( n - 4 = 0 \) → \( n = 4 \)
2. \( n + 3 = 0 \) → \( n = -3 \) (not a valid quantum number)

Thus, the only valid solution is:

\[ n = 4 \]

### Conclusion

The principal quantum number of the excited state is \( n = 4 \).

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