The change in velocity when electron jumps from the first orbit to the second orbit is

Options

A. Half its original velocity
B. Twice its original velocity
C. One fourth its original velocity
D. Equal to its original velocity

Answer

To find the change in velocity when an electron jumps from the first orbit to the second orbit, we can follow these steps:

### Step 1: Understand the Energy Levels
The total energy (E_total) of an electron in an orbit is the sum of its potential energy (PE) and kinetic energy (KE). For an electron revolving around a nucleus, the potential energy is given by:
\[ PE = -\frac{ze^2}{r} \]
And the kinetic energy is given by:
\[ KE = \frac{ze^2}{2r} \]

### Step 2: Total Energy in Different Orbits
For a hydrogen atom, the total energy for the nth orbit can be expressed as:
\[ E_n = -\frac{z e^2}{2r} \]
where \( r \) is the radius of the orbit.

### Step 3: Calculate Energies for n=1 and n=2
For the first orbit (n=1):
\[ E_1 = -\frac{z e^2}{2r_1} \]
For the second orbit (n=2):
\[ E_2 = -\frac{z e^2}{2r_2} \]

### Step 4: Relate the Radii of the Orbits
The radius of the orbits can be related to the principal quantum number (n):
\[ r_n = n^2 r_0 \]
where \( r_0 \) is a constant (the Bohr radius). Thus:
- For n=1: \( r_1 = r_0 \)
- For n=2: \( r_2 = 4r_0 \)

### Step 5: Calculate the Kinetic Energies
The kinetic energy for the first orbit (n=1):
\[ KE_1 = \frac{ze^2}{2r_1} = \frac{ze^2}{2r_0} \]
The kinetic energy for the second orbit (n=2):
\[ KE_2 = \frac{ze^2}{2r_2} = \frac{ze^2}{8r_0} \]

### Step 6: Relate the Kinetic Energies
Since the mass of the electron remains constant, we can relate the velocities using the kinetic energy formula:
\[ KE = \frac{1}{2} mv^2 \]
Thus:
\[ KE_1 = \frac{1}{2} mv_1^2 \]
\[ KE_2 = \frac{1}{2} mv_2^2 \]

### Step 7: Set Up the Ratio of Kinetic Energies
From the kinetic energy expressions:
\[ \frac{KE_1}{KE_2} = \frac{v_1^2}{v_2^2} \]
Substituting the values:
\[ \frac{\frac{ze^2}{2r_0}}{\frac{ze^2}{8r_0}} = \frac{v_1^2}{v_2^2} \]
This simplifies to:
\[ \frac{4}{1} = \frac{v_1^2}{v_2^2} \]

### Step 8: Solve for the Velocities
Taking the square root:
\[ \frac{v_1}{v_2} = 2 \]
Thus:
\[ v_2 = \frac{v_1}{2} \]

### Step 9: Determine the Change in Velocity
The change in velocity when the electron jumps from the first orbit to the second orbit is:
\[ \Delta v = v_2 - v_1 = \frac{v_1}{2} - v_1 = -\frac{v_1}{2} \]

### Conclusion
The change in velocity is half of its original velocity, indicating that the electron moves slower in the second orbit compared to the first.

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