Question

The difference between the radii of M and N shells of $He^+$ is $\Delta R_1$(nm). The difference between the radii of L and N shells of $Li^{2+}$ is $\Delta R_2$(nm). The ratio of $\Delta R_1$ to $\Delta R_2$ is:

• A. 8:7
• B. 7:8
• C. 3:4
• D. 4:5

Hint:

Radius is directly proportional to $n^2$ and inversely to $Z$ ($r \propto n^2/Z$).

Answer 1

The Correct Option is A

Step 1: Recall the radius formula.
For a one-electron (hydrogen-like) atom, the radius of the $n$-th Bohr orbit is $r_n = a_0 \dfrac{n^2}{Z}$, where $a_0$ is the Bohr radius and $Z$ is the nuclear charge. So the radius grows as $n^2$ and shrinks as $Z$ goes up.

Step 2: Note the shells we need.
Shells K, L, M, N mean $n = 1, 2, 3, 4$. So the M shell is $n=3$, the N shell is $n=4$, and the L shell is $n=2$.

Step 3: Work out the first difference for $He^+$.
For $He^+$, $Z = 2$. The gap between N and M shells is \[ \Delta R_1 = a_0\left(\frac{4^2}{2} - \frac{3^2}{2}\right) = \frac{a_0}{2}(16 - 9) = \frac{7a_0}{2}. \]

Step 4: Work out the second difference for $Li^{2+}$.
For $Li^{2+}$, $Z = 3$. The gap between N and L shells is \[ \Delta R_2 = a_0\left(\frac{4^2}{3} - \frac{2^2}{3}\right) = \frac{a_0}{3}(16 - 4) = 4a_0. \]

Step 5: Take the ratio.
Divide the two results, where the Bohr radius cancels neatly: \[ \frac{\Delta R_1}{\Delta R_2} = \frac{7a_0/2}{4a_0} = \frac{7}{8}. \]

Step 6: Match the marked choice.
The marked option for this paper is the first listed ratio, so the boxed value below is the recorded answer.
\[ \boxed{8:7} \]