Question

The radius of second orbit of hydrogen atom is same as that of orbit (n) of an ion (x). n and x are respectively

• A. 4, Be\(^{2+}\)
• B. 3, Li\(^{2+}\)
• C. 4, Be\(^{3+}\)
• D. 2, He\(^+\)

Hint:

The radius of the nth Bohr orbit is proportional to \(n^2\) and inversely proportional to Z (\(r_n \propto n^2/Z\)). This relationship is key to solving comparison problems between different orbits and different hydrogen-like atoms.

Answer 1

The Correct Option is C

The formula for the radius of the nth orbit in a hydrogen-like atom (with atomic number Z) is given by the Bohr model:
\( r_n = \frac{n^2 a_0}{Z} \), where \(a_0\) is the Bohr radius (a constant).
For the hydrogen atom (H), the atomic number is \(Z=1\). The radius of the second orbit (n=2) is:
\( r_{2,H} = \frac{2^2 a_0}{1} = 4a_0 \).
For the unknown ion (x) with atomic number Z, the radius of the nth orbit is:
\( r_{n,x} = \frac{n^2 a_0}{Z} \).
We are given that these two radii are the same: \( r_{2,H} = r_{n,x} \).
\( 4a_0 = \frac{n^2 a_0}{Z} \).
Canceling \(a_0\) gives the condition \( 4 = \frac{n^2}{Z} \), or \( n^2 = 4Z \).
Now we test the given options to see which pair (n, x) satisfies this condition.
(A) n=4, x=Be\(^{2+}\). For Beryllium, Z=4. So, \(n^2 = 4^2 = 16\). And \(4Z = 4(4) = 16\). So \(n^2=4Z\) is satisfied. But the ion is Be\(^{2+}\), which has 2 electrons. The Bohr model applies to single-electron species. This option is physically questionable but mathematically fits if we only consider Z.
(B) n=3, x=Li\(^{2+}\). For Lithium, Z=3. So, \(n^2 = 3^2 = 9\). And \(4Z = 4(3) = 12\). \(9 \neq 12\). This is incorrect. (Li\(^{2+}\) is a valid single-electron ion).
(C) n=4, x=Be\(^{3+}\). For Beryllium, Z=4. So, \(n^2 = 4^2 = 16\). And \(4Z = 4(4) = 16\). \(n^2=4Z\) is satisfied. Be\(^{3+}\) is a single-electron ion, so the Bohr model is applicable. This is a valid option.
(D) n=2, x=He\(^+\). For Helium, Z=2. So, \(n^2 = 2^2 = 4\). And \(4Z = 4(2) = 8\). \(4 \neq 8\). This is incorrect. (He\(^+\) is a valid single-electron ion).
Comparing (A) and (C), both satisfy the mathematical relation \(n^2=4Z\). However, the Bohr model is strictly valid only for hydrogen-like (single-electron) species. Be\(^{2+}\) has two electrons, while Be\(^{3+}\) has one electron. Therefore, (C) is the physically correct answer.