The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV . The magnitudes energies from three lowest orbits of the hydrogen are

Options

A. 13.6, 6.8 , 8.4 eV
B. 13.6 , 10.2 , 3.4 eV
C. 13.6 , 27.2 , 40.8 eV
D. 13.6 ,3.4, 1.5 eV

Answer

To find the binding energies of the first three orbits of the hydrogen atom, we can use the formula for the energy of an electron in the nth orbit of a hydrogen atom, which is given by:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

where:
- \(E_n\) is the energy of the nth orbit,
- \(13.6 \, \text{eV}\) is the binding energy of the electron in the lowest orbit (ground state),
- \(n\) is the principal quantum number (1 for the first orbit, 2 for the second orbit, and so on).

### Step-by-Step Solution:

1. **Calculate the energy for the first orbit (n=1)**:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

2. **Calculate the energy for the second orbit (n=2)**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

3. **Calculate the energy for the third orbit (n=3)**:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

4. **Summarize the magnitudes of the energies**:
- For the first orbit: \( |E_1| = 13.6 \, \text{eV} \)
- For the second orbit: \( |E_2| = 3.4 \, \text{eV} \)
- For the third orbit: \( |E_3| \approx 1.51 \, \text{eV} \)

### Final Answer:
The magnitudes of the energies from the three lowest orbits of the hydrogen atom are:
- 13.6 eV (first orbit)
- 3.4 eV (second orbit)
- 1.51 eV (third orbit)