If the velocity of an elelctron moving in first bohr 's orbit of hydrogen has to be reduced to `7.272 xx 10^(5) ms^(-1)` it should be shifted by

Options

A. 4.23 A away from nucleus
B. 4.23 A towards nucleus
C. 4.76 A away from nucleus
D. 4.76 A towards nucleus

Answer

To solve the problem of how much the electron's orbit needs to shift when its velocity is reduced to \(7.272 \times 10^5 \, \text{m/s}\), we can follow these steps:

### Step 1: Write down the given velocity
The velocity of the electron is given as:
\[ V = 7.272 \times 10^5 \, \text{m/s} \]

### Step 2: Use the formula for the velocity of an electron in Bohr's model
The formula for the velocity of an electron in the nth orbit of a hydrogen atom (where \(Z = 1\)) is:
\[ V = \frac{2.18 \times 10^6 \cdot Z}{n} \]
For hydrogen, \(Z = 1\), so the formula simplifies to:
\[ V = \frac{2.18 \times 10^6}{n} \]

### Step 3: Substitute the given velocity into the formula
Substituting the given velocity into the equation:
\[ 7.272 \times 10^5 = \frac{2.18 \times 10^6}{n} \]

### Step 4: Solve for \(n\)
Rearranging the equation to solve for \(n\):
\[ n = \frac{2.18 \times 10^6}{7.272 \times 10^5} \]
Calculating \(n\):
\[ n \approx 3 \]

### Step 5: Calculate the radius for the nth orbit
Using the formula for the radius of the nth orbit in Bohr's model:
\[ r = \frac{0.529 \, n^2}{Z} \]
Substituting \(n = 3\) and \(Z = 1\):
\[ r = 0.529 \times 3^2 = 0.529 \times 9 = 4.761 \, \text{Å} \]

### Step 6: Calculate the shift in radius
The radius of the first orbit (for \(n = 1\)) is:
\[ r_1 = 0.529 \, \text{Å} \]
Now, we find the shift:
\[ \text{Shift} = r - r_1 = 4.761 \, \text{Å} - 0.529 \, \text{Å} = 4.232 \, \text{Å} \]

### Step 7: Determine the direction of the shift
Since the new radius \(4.761 \, \text{Å}\) is greater than the original radius \(0.529 \, \text{Å}\), the electron must shift **away from the nucleus**.

### Final Answer
The electron should be shifted by approximately **4.23 Å away from the nucleus**.

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