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. `4 :3 `
B. `3:4`
C. `2:1`
D. `1:2`

Explanation

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

### Step 1: Determine \( n_1 \)
The number of spectral lines emitted by electrons in a given energy level \( n \) can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]
Given that for \( n = n_1 \), the number of spectral lines is 3, we can set up the equation:
\[ \frac{n_1(n_1 - 1)}{2} = 3 \]
Multiplying both sides by 2 gives:
\[ n_1(n_1 - 1) = 6 \]
Now, we can solve for \( n_1 \):
\[ n_1^2 - n_1 - 6 = 0 \]
Factoring the quadratic equation:
\[ (n_1 - 3)(n_1 + 2) = 0 \]
This gives us \( n_1 = 3 \) (since \( n \) cannot be negative).

### Step 2: Determine \( n_2 \)
For \( n = n_2 \), the number of spectral lines is 6. Using the same formula:
\[ \frac{n_2(n_2 - 1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n_2(n_2 - 1) = 12 \]
Now, we can solve for \( n_2 \):
\[ n_2^2 - n_2 - 12 = 0 \]
Factoring the quadratic equation:
\[ (n_2 - 4)(n_2 + 3) = 0 \]
This gives us \( n_2 = 4 \) (again, \( n \) cannot be negative).

### Step 3: Find the ratio of orbital speeds
The orbital speed \( v \) of electrons in a hydrogen-like atom is given by the formula:
\[ v \propto \frac{Z}{n} \]
where \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)). The ratio of the speeds in the two orbits is:
\[ \frac{v_{n_1}}{v_{n_2}} = \frac{n_2}{n_1} \]
Substituting the values we found:
\[ \frac{v_{n_1}}{v_{n_2}} = \frac{4}{3} \]

### Final Answer
The ratio of the orbital speeds of the electrons in the two orbits is:
\[ \frac{v_{n_1}}{v_{n_2}} = \frac{4}{3} \]