The hydrogen is excited and its Bohr radius changes from `5.3 xx 10^(-11)` m to `47.5 xx 10^(-11)` m. Calculate the principal quantum number and also find the total energy of atom.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Bohr Radius Formula
The radius of the nth orbit of a hydrogen atom is given by the formula:
\[ r_n = 0.53 \times n^2 \, \text{Å} \]
where \( r_n \) is the radius in angstroms, and \( n \) is the principal quantum number.

### Step 2: Calculate the Principal Quantum Number for the Initial Radius
Given the initial radius \( r_1 = 5.3 \times 10^{-11} \, \text{m} \) (which is equivalent to \( 0.53 \, \text{Å} \)):
Using the formula:
\[ 0.53 = 0.53 \times n_1^2 \]
Dividing both sides by \( 0.53 \):
\[ 1 = n_1^2 \]
Taking the square root:
\[ n_1 = 1 \]

### Step 3: Calculate the Principal Quantum Number for the Final Radius
Given the final radius \( r_2 = 47.5 \times 10^{-11} \, \text{m} \) (which is equivalent to \( 4.75 \, \text{Å} \)):
Using the formula:
\[ 4.75 = 0.53 \times n_2^2 \]
Dividing both sides by \( 0.53 \):
\[ n_2^2 = \frac{4.75}{0.53} \]
Calculating the right side:
\[ n_2^2 \approx 8.96 \]
Taking the square root:
\[ n_2 \approx 3 \]

### Step 4: Conclusion on Principal Quantum Number
The hydrogen atom transitions from the first orbit (n=1) to the third orbit (n=3). Thus, the principal quantum number after excitation is:
\[ n = 3 \]

### Step 5: Calculate the Total Energy of the Atom
The total energy of the electron in the nth orbit is given by:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
For hydrogen, \( Z = 1 \):
\[ E_3 = -\frac{13.6 \times 1^2}{3^2} \]
Calculating:
\[ E_3 = -\frac{13.6}{9} \]
\[ E_3 \approx -1.51 \, \text{eV} \]

### Final Answer
The principal quantum number after excitation is \( n = 3 \), and the total energy of the hydrogen atom in this state is approximately \( -1.51 \, \text{eV} \).

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