Ration of time revolution of electron in second excited state of `He^(+)` and first state of `H` is `(3^(x))/(2^(y))`.
`(x + y)` is

Options

A. `9`
B. `7`
C. `8`
D. none of these

Explanation

To solve the problem of finding the ratio of the time of revolution of an electron in the second excited state of \( He^+ \) and the first state of \( H \), we will follow these steps:

### Step 1: Identify the states
- The second excited state of \( He^+ \) corresponds to \( n = 3 \).
- The first state of \( H \) corresponds to \( n = 2 \).

### Step 2: Use the formula for frequency
The frequency (\( \nu \)) of an electron in a hydrogen-like atom is given by the formula:
\[ \nu \propto \frac{Z^2}{n^3} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 3: Calculate the frequency for \( He^+ \)
For \( He^+ \):
- \( Z = 2 \) (since Helium has an atomic number of 2)
- \( n = 3 \) (second excited state)

Thus,
\[ \nu_{He^+} \propto \frac{2^2}{3^3} = \frac{4}{27} \]

### Step 4: Calculate the frequency for \( H \)
For \( H \):
- \( Z = 1 \) (Hydrogen has an atomic number of 1)
- \( n = 2 \) (first state)

Thus,
\[ \nu_H \propto \frac{1^2}{2^3} = \frac{1}{8} \]

### Step 5: Find the ratio of frequencies
Now we can find the ratio of the frequencies:
\[ \frac{\nu_{He^+}}{\nu_H} = \frac{\frac{4}{27}}{\frac{1}{8}} = \frac{4}{27} \times \frac{8}{1} = \frac{32}{27} \]

### Step 6: Express the ratio in terms of powers of 2 and 3
We can express \( \frac{32}{27} \) in terms of powers of 2 and 3:
\[ 32 = 2^5 \quad \text{and} \quad 27 = 3^3 \]
Thus,
\[ \frac{\nu_{He^+}}{\nu_H} = \frac{2^5}{3^3} \]

### Step 7: Identify \( x \) and \( y \)
From the expression \( \frac{2^x}{3^y} \), we can see that:
- \( x = 5 \)
- \( y = 3 \)

### Step 8: Calculate \( x + y \)
Now, we can find \( x + y \):
\[ x + y = 5 + 3 = 8 \]

### Final Answer
Thus, the value of \( x + y \) is \( 8 \).

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