An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transition. An which there can be a maximum of three transition. Find ratio of the velocities of the electron in these two orbits.

Options

A. `1/2`
B. `2/3`
C. `5/4`
D. `3/4`

Answer

To solve the problem, we need to find the ratio of the velocities of an electron in two different orbits of a hydrogen atom, given the maximum number of transitions possible from each orbit.

### Step 1: Understanding the Number of Transitions
The maximum number of spectral lines (transitions) that can occur when an electron transitions between energy levels is given by the formula:
\[ N = \frac{n(n-1)}{2} \]
where \( N \) is the number of transitions and \( n \) is the principal quantum number of the energy level.

### Step 2: Finding the Principal Quantum Number for the First Orbit
For the first orbit where there are a maximum of 6 transitions:
\[ N_1 = 6 = \frac{n_1(n_1 - 1)}{2} \]
Multiplying both sides by 2 gives:
\[ 12 = n_1(n_1 - 1) \]
This simplifies to:
\[ n_1^2 - n_1 - 12 = 0 \]
Factoring or using the quadratic formula, we find:
\[ n_1 = 4 \]

### Step 3: Finding the Principal Quantum Number for the Second Orbit
For the second orbit where there are a maximum of 3 transitions:
\[ N_2 = 3 = \frac{n_2(n_2 - 1)}{2} \]
Multiplying both sides by 2 gives:
\[ 6 = n_2(n_2 - 1) \]
This simplifies to:
\[ n_2^2 - n_2 - 6 = 0 \]
Factoring or using the quadratic formula, we find:
\[ n_2 = 3 \]

### Step 4: Finding the Ratio of Velocities
The velocity of an electron in a hydrogen atom is proportional to the principal quantum number:
\[ v \propto \frac{1}{n} \]
Thus, the ratio of the velocities of the electron in the two orbits is:
\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4} \]

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