The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from three lowest energy orbits of hydrogen atom respectively are

Options

A. `13.6 6.8 , 8.4`
B. `13.6 , 10.2 ,3.4`
C. `13.6 , 27.2 , 40.8`
D. `13.6 ,3.4,1.51`

Answer

To solve the problem of finding the energies required to remove an electron from the three lowest energy orbits of a hydrogen atom, we can use the formula for the energy levels of the hydrogen atom:

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

Where:
- \( E_n \) is the energy of the electron in the nth orbit,
- \( n \) is the principal quantum number,
- The negative sign indicates that the electron is bound to the nucleus.

### Step-by-step Solution:

1. **Calculate the Energy for the First Orbit (n=1)**:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
The energy required to remove the electron (ionization energy) is the absolute value:
\[ \text{Ionization Energy from n=1} = 13.6 \, \text{eV} \]

2. **Calculate the Energy for the Second Orbit (n=2)**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
The energy required to remove the electron from this orbit is:
\[ \text{Ionization Energy from n=2} = 3.4 \, \text{eV} \]

3. **Calculate the Energy for the Third Orbit (n=3)**:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]
The energy required to remove the electron from this orbit is:
\[ \text{Ionization Energy from n=3} \approx 1.51 \, \text{eV} \]

### Final Results:
The energies required to remove an electron from the three lowest energy orbits of the hydrogen atom are:
- For n=1: **13.6 eV**
- For n=2: **3.4 eV**
- For n=3: **1.51 eV**

### Summary:
The energies required to remove an electron from the three lowest energy orbits of hydrogen atom respectively are:
- **13.6 eV, 3.4 eV, 1.51 eV**