Electrons in a certain energy level `n=n_(1)` can emit 3 spectral lines. When they are in another energy level, `n=n_(2)`, they can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

Options

A. (a)`4:3`
B. (b)`3:4`
C. (c)`2:1`
D. (d)`1:2`

Answer

To solve the problem, we need to find the ratio of the orbital speeds of electrons in two different energy levels, \( n_1 \) and \( n_2 \), based on the number of spectral lines emitted from those energy levels.

### Step-by-Step Solution:

1. **Understanding the Number of Spectral Lines**:
The number of spectral lines \( N \) that can be emitted when electrons transition between energy levels is given by the formula:
\[ N = \frac{n(n-1)}{2} \]
where \( n \) is the principal quantum number of the energy level.

2. **Applying the Formula for \( n_1 \)**:
For the first energy level \( n_1 \), we know that it emits 3 spectral lines:
\[ \frac{n_1(n_1-1)}{2} = 3 \]
Multiplying both sides by 2 gives:
\[ n_1(n_1-1) = 6 \]
Rearranging this gives us the quadratic equation:
\[ n_1^2 - n_1 - 6 = 0 \]

3. **Solving the Quadratic Equation for \( n_1 \)**:
We can factor this equation:
\[ (n_1 - 3)(n_1 + 2) = 0 \]
Thus, \( n_1 = 3 \) (we discard \( n_1 = -2 \) since \( n \) must be positive).

4. **Applying the Formula for \( n_2 \)**:
For the second energy level \( n_2 \), it emits 6 spectral lines:
\[ \frac{n_2(n_2-1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n_2(n_2-1) = 12 \]
Rearranging this gives us the quadratic equation:
\[ n_2^2 - n_2 - 12 = 0 \]

5. **Solving the Quadratic Equation for \( n_2 \)**:
We can factor this equation:
\[ (n_2 - 4)(n_2 + 3) = 0 \]
Thus, \( n_2 = 4 \) (we discard \( n_2 = -3 \) since \( n \) must be positive).

6. **Finding the Ratio of Orbital Speeds**:
The speed of the electron in an orbit is inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \]
Therefore, the ratio of the speeds \( \frac{v_1}{v_2} \) can be expressed as:
\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]
Substituting the values we found:
\[ \frac{v_1}{v_2} = \frac{4}{3} \]

### Final Answer:
The ratio of the orbital speeds of the electrons in the two orbits is:
\[ \frac{v_1}{v_2} = \frac{4}{3} \]