A sample of hydrogen gas is excited by means of a monochromatic radiation. In the subsequent emission spectrum, `10` differenet wavelengths are obtained, all of which have enrgies greater than or equal to the energy of the abosrbed radiation. Find the intitial quantum number of the state (before abosrbing radiation).

Explanation

To solve the problem, we need to determine the initial quantum number of the hydrogen atom before it absorbs radiation that leads to the emission of 10 different wavelengths. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between quantum numbers and transitions
In a hydrogen atom, the energy levels are quantized and can be described by principal quantum numbers (n). When an electron transitions from a higher energy level (n_initial) to a lower energy level (n_final), it emits a photon with a specific wavelength.

### Step 2: Determine the number of transitions
The number of different wavelengths (or spectral lines) emitted can be calculated using the formula for combinations:
\[ \text{Number of transitions} = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels involved in the transitions.

### Step 3: Set up the equation
Given that there are 10 different wavelengths, we can set up the equation:
\[ \frac{n(n-1)}{2} = 10 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 20 \]

### Step 4: Solve the quadratic equation
Rearranging gives us:
\[ n^2 - n - 20 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = -20 \):
\[ n = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2} \]
This gives us two potential solutions for \( n \):
\[ n = 5 \quad \text{or} \quad n = -4 \]
Since \( n \) must be a positive integer, we have \( n = 5 \).

### Step 5: Determine the initial quantum number
The problem states that all emitted wavelengths have energies greater than or equal to the energy of the absorbed radiation. The absorbed energy corresponds to the transition from the initial quantum number \( n_{initial} \) to a higher state.

Since we have determined \( n = 5 \) as the highest energy level involved in the transitions, the initial quantum number must be one less than this, because the transitions can only occur to lower energy levels. Therefore:
\[ n_{initial} = 5 - 1 = 4 \]

### Final Answer
The initial quantum number of the state before absorbing the radiation is:
\[ \boxed{4} \]