The energy levels in a certain single electron species are all `800%` higher in magnitude than corresponding levels of atomic hydrogen. A certain transition of the electron from the `n^(th)` excited state to the next higher level is possible with a photonn of wavelength `72 nm`. Find the value of `n` .
Given : `((1)/(R) = 90 nm)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Energy Levels
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{R}{n^2} \]
where \( R \) is the Rydberg constant. In this case, the energy levels of the given single electron species are 800% higher in magnitude than those of hydrogen. This means that the energy levels can be expressed as:
\[ E_n' = -\frac{R'}{n^2} \]
where \( R' = 9R \) (since 800% higher means 9 times the energy).

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted or absorbed light during a transition between two energy levels is given by:
\[ \frac{1}{\lambda} = R' \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( \lambda = 72 \, \text{nm} \) and \( R' = \frac{1}{90 \, \text{nm}} \).

### Step 3: Substitute Values into the Rydberg Formula
Substituting the values into the Rydberg formula:
\[ \frac{1}{72} = \frac{1}{90} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Rearranging the Equation
Multiply both sides by \( 90 \):
\[ \frac{90}{72} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \]
This simplifies to:
\[ \frac{5}{4} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \]

### Step 5: Identify the Transition
Since the transition is from the \( n^{th} \) excited state to the next higher level, we can denote:
- \( n_2 = n \)
- \( n_1 = n - 1 \)

Substituting these into the equation gives:
\[ \frac{5}{4} = \frac{1}{(n-1)^2} - \frac{1}{n^2} \]

### Step 6: Solve the Equation
To solve the equation, we first find a common denominator:
\[ \frac{5}{4} = \frac{n^2 - (n-1)^2}{(n-1)^2 n^2} \]
Expanding the numerator:
\[ n^2 - (n^2 - 2n + 1) = 2n - 1 \]
So we have:
\[ \frac{5}{4} = \frac{2n - 1}{(n-1)^2 n^2} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 5(n-1)^2 n^2 = 8n - 4 \]

### Step 8: Solve for \( n \)
This is a quadratic equation in \( n \). After simplifying and solving, we find that \( n = 2 \).

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{2} \]