The ratio between acceleration of the electron in singlely ionized helium atom and doublely ionized lithium atom (both in ground state) is (x/27). Find value of x.

Answer

To find the ratio between the acceleration of the electron in a singly ionized helium atom (He\(^+\)) and a doubly ionized lithium atom (Li\(^{2+}\)), we will follow these steps:

### Step 1: Understand the formula for centripetal acceleration
The centripetal acceleration \(a\) of an electron revolving around the nucleus can be expressed as:
\[ a = \frac{V^2}{r} \]
where \(V\) is the velocity of the electron and \(r\) is the radius of the orbit.

### Step 2: Determine the velocity of the electron
The velocity \(V\) of the electron in an atom is given by:
\[ V \propto \frac{Z}{n} \]
where \(Z\) is the atomic number and \(n\) is the principal quantum number. For both helium and lithium, we will consider the ground state, where \(n = 1\).

### Step 3: Determine the radius of the electron's orbit
The radius \(r\) of the electron's orbit is given by:
\[ r \propto \frac{n^2}{Z} \]
Again, for the ground state, \(n = 1\).

### Step 4: Combine the expressions for acceleration
Substituting the expressions for \(V\) and \(r\) into the centripetal acceleration formula:
\[ a \propto \frac{(Z/n)^2}{n^2/Z} = \frac{Z^2}{Z} = Z^3 \]
Thus, we can conclude that the acceleration is proportional to \(Z^3\).

### Step 5: Calculate the accelerations for He\(^+\) and Li\(^{2+}\)
For singly ionized helium (He\(^+\)):
- \(Z = 2\)

For doubly ionized lithium (Li\(^{2+}\)):
- \(Z = 3\)

### Step 6: Find the ratio of accelerations
Now we can find the ratio of the accelerations:
\[ \frac{a_{He^+}}{a_{Li^{2+}}} = \frac{Z_{He}^3}{Z_{Li}^3} = \frac{2^3}{3^3} = \frac{8}{27} \]

### Step 7: Identify the value of \(x\)
From the ratio \(\frac{a_{He^+}}{a_{Li^{2+}}} = \frac{x}{27}\), we can see that:
\[ \frac{8}{27} = \frac{x}{27} \]
Thus, \(x = 8\).

### Final Answer
The value of \(x\) is \(8\).

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