The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio 1:4. The energy difference between them may be

Options

A. 0.85 eV
B. 10.2 eV
C. 3.40 eV
D. 13.6 eV

Answer

To solve the problem, we need to find the energy difference between two Bohr orbits of the hydrogen atom, given that the radii of these orbits are in the ratio of 1:4.

### Step-by-Step Solution:

1. **Identify the relationship between radius and principal quantum number (n)**:
The radius of the nth orbit in a hydrogen atom is given by the formula:
\[ R_n = n^2 \cdot R_1 \]
where \( R_1 \) is the radius of the first orbit (n=1).

2. **Set up the ratio of the radii**:
According to the problem, the ratio of the radii of two orbits is 1:4. This means:
\[ \frac{R_{n'}}{R_n} = \frac{1}{4} \]
Using the formula for the radius, we can express this as:
\[ \frac{n'^2 \cdot R_1}{n^2 \cdot R_1} = \frac{1}{4} \]
Simplifying gives:
\[ \frac{n'^2}{n^2} = \frac{1}{4} \]

3. **Determine the values of n and n'**:
From the ratio \( \frac{n'^2}{n^2} = \frac{1}{4} \), we can deduce:
\[ n' = 2 \quad \text{and} \quad n = 1 \]

4. **Calculate the energy difference**:
The energy of an electron in the nth orbit of hydrogen is given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Therefore, the energies for n=1 and n'=2 are:
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

5. **Find the energy difference (ΔE)**:
The energy difference between the two levels is:
\[ \Delta E = E_{n'} - E_n = E_2 - E_1 \]
Substituting the values:
\[ \Delta E = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

### Final Answer:
The energy difference between the two orbits is \( \Delta E = 10.2 \, \text{eV} \).