Question

Find the radius of the orbit corresponding to the 4th excited state in Li++. (a0 is the radius of first orbit in H-atom)

• A. \(\frac{25}{3}\, a_0\)
• B. \(\frac{16}{3}\, a_0\)
• C. \(25\, a_0\)
• D. \(12\, a_0\)

Answer 1

The Correct Option is A

To solve the problem of finding the radius of the orbit corresponding to the 4th excited state in the Li⁺⁺ ion, we first need to understand the concept of orbit radii in hydrogen-like ions.

The radius of an orbit in a hydrogen-like atom is given by the formula:

\(r_n = \frac{n^2 a_0}{Z}\)

Here:

• \(r_n\) is the radius of the nth orbit.
• \(n\) is the principal quantum number (energy level).
• \(a_0\) is the Bohr radius (radius of the hydrogen atom's first orbit).
• \(Z\) is the atomic number of the nucleus.

For lithium ion Li⁺⁺ (similar to a hydrogen-like atom but with a higher atomic number), \(Z = 3\).

The 4th excited state corresponds to an energy level of \(n = 5\) (since the ground state \((n=1)\) plus 4 excitations gives \(n=5\)).

Substituting the values into the formula:

\(r_5 = \frac{5^2 \cdot a_0}{3} = \frac{25 \cdot a_0}{3}\)

Therefore, the radius of the orbit corresponding to the 4th excited state in the Li⁺⁺ ion is \(\frac{25}{3}\, a_0\).

The correct answer is:

\(\frac{25}{3}\, a_0\)