if speed of electron is ground state energy level is `2.2xx10^(6)ms^(-1)`, then its speed in fourth excited state will be

Options

A. `6.8xx10^(6)ms^(-1)`
B. `8.8xx10^(5)ms^(-1)`
C. `5.5xx10^(5)ms^(-1)`
D. `5.5xx10^(6)ms^(-1)`

Explanation

To find the speed of an electron in the fourth excited state given its speed in the ground state, we can use the relationship derived from Bohr's model of the atom. Here’s a step-by-step solution:

### Step 1: Understand the relationship between speed and quantum number
According to Bohr's model, the speed of an electron in an orbit is inversely proportional to the principal quantum number \( n \). This can be expressed as:
\[ V \propto \frac{1}{n} \]
where \( V \) is the speed of the electron and \( n \) is the principal quantum number.

### Step 2: Set up the ratio of speeds
Let \( V_1 \) be the speed of the electron in the ground state (where \( n = 1 \)) and \( V_4 \) be the speed in the fourth excited state (where \( n = 4 \)). The relationship can be written as:
\[ \frac{V_1}{V_4} = \frac{n_4}{n_1} \]
Substituting the values of \( n_4 = 4 \) and \( n_1 = 1 \), we have:
\[ \frac{V_1}{V_4} = \frac{4}{1} \]

### Step 3: Substitute the known value
We know from the problem that \( V_1 = 2.2 \times 10^6 \, \text{m/s} \). Substituting this into the equation gives:
\[ \frac{2.2 \times 10^6}{V_4} = 4 \]

### Step 4: Solve for \( V_4 \)
To find \( V_4 \), we can rearrange the equation:
\[ V_4 = \frac{2.2 \times 10^6}{4} \]
Calculating this gives:
\[ V_4 = 0.55 \times 10^6 = 5.5 \times 10^5 \, \text{m/s} \]

### Conclusion
Thus, the speed of the electron in the fourth excited state is:
\[ V_4 = 5.5 \times 10^5 \, \text{m/s} \]

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