The ratio of the energy required to remove an electron from the first three Bohr's orbit of Hydrogen atom is

Options

A. `3 : 2 : 1`
B. `9 : 4 :1`
C. `36 : 9: 4`
D. `1: 4: 9`

Answer

To find the ratio of the energy required to remove an electron from the first three Bohr orbits of a hydrogen atom, we can use the formula derived from Bohr's model of the atom. The energy of an electron in the nth orbit is given by:

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

where \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)) and \( n \) is the principal quantum number (1, 2, 3, ...).

### Step-by-Step Solution:

1. **Calculate \( E_1 \)**:
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \, \text{eV} \]

2. **Calculate \( E_2 \)**:
\[ E_2 = -\frac{13.6 \times 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

3. **Calculate \( E_3 \)**:
\[ E_3 = -\frac{13.6 \times 1^2}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]

4. **Set up the ratio**:
The ratio of the energies \( E_1 : E_2 : E_3 \) is:
\[ E_1 : E_2 : E_3 = -13.6 : -3.4 : -1.51 \]

5. **Remove the negative signs**:
\[ E_1 : E_2 : E_3 = 13.6 : 3.4 : 1.51 \]

6. **Convert to a common ratio**:
To simplify the ratio, we can express each term with a common denominator. First, we can convert the values to fractions:
\[ E_1 = 13.6, \quad E_2 = 3.4 = \frac{13.6}{4}, \quad E_3 = \frac{13.6}{9} \]

7. **Express the ratio**:
\[ E_1 : E_2 : E_3 = 1 : \frac{1}{4} : \frac{1}{9} \]

8. **Find the least common multiple (LCM)**:
The LCM of 4 and 9 is 36. Multiply each term by 36:
\[ E_1 : E_2 : E_3 = 36 \times 1 : 36 \times \frac{1}{4} : 36 \times \frac{1}{9} \]
This gives:
\[ E_1 : E_2 : E_3 = 36 : 9 : 4 \]

### Final Ratio:
The final ratio of the energy required to remove an electron from the first three Bohr orbits of a hydrogen atom is:
\[ \text{Ratio} = 36 : 9 : 4 \]