The energy, radius and velocity of the electron in the hydrogen atom in the ground state are `-13.6 eV, 0.53 Å` and `2.188 xx 10^(8) m s^(-1)` respectively.
If the electron absorbs `12.1 eV` of energy, it will jump to the orbit

Options

A. 2nd
B. 3rd
C. 4th
D. 5th

Explanation

To solve the problem, we need to determine the orbit to which the electron in a hydrogen atom will jump after absorbing 12.1 eV of energy. We will follow these steps:

### Step 1: Understand the Ground State Energy
The energy of the electron in the ground state (n=1) of the hydrogen atom is given as:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Calculate the Total Energy After Absorption
When the electron absorbs 12.1 eV of energy, the new energy (E_n) of the electron can be calculated as:
\[ E_n = E_1 + \text{Energy Absorbed} \]
\[ E_n = -13.6 \, \text{eV} + 12.1 \, \text{eV} \]
\[ E_n = -1.5 \, \text{eV} \]

### Step 3: Use the Energy Formula for Hydrogen Atom
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

### Step 4: Set Up the Equation
We can now set up the equation using the energy we calculated:
\[ -1.5 = -\frac{13.6}{n^2} \]

### Step 5: Solve for n^2
Removing the negative signs and rearranging gives us:
\[ 1.5 = \frac{13.6}{n^2} \]
\[ n^2 = \frac{13.6}{1.5} \]

### Step 6: Calculate n^2
Now, we perform the calculation:
\[ n^2 = \frac{13.6}{1.5} \approx 9.0667 \]

### Step 7: Determine n
Taking the square root gives us:
\[ n \approx 3 \]

### Conclusion
Thus, the electron will jump to the orbit with principal quantum number:
\[ n = 3 \]