Step 1: Recall Bohr's quantization condition.
For an electron in the $ n $-th orbit of a hydrogen-like atom, Bohr's condition states that the circumference equals an integer multiple of the de Broglie wavelength: \[ 2\pi r_n = n\lambda_n \implies \lambda_n = \frac{2\pi r_n}{n} \]
Step 2: Use Bohr's orbital radius formula.
The radius of the $ n $-th orbit for a hydrogen-like atom with atomic number $ Z $ is: \[ r_n = \frac{n^2 a_0}{Z} \] where $ a_0 $ is the Bohr radius. Substituting into the wavelength expression: \[ \lambda_n = \frac{2\pi}{n}\cdot\frac{n^2 a_0}{Z} = \frac{2\pi n a_0}{Z} \] Therefore $ \lambda \propto \frac{n}{Z} $.
Step 3: Calculate wavelength for hydrogen ground state.
For hydrogen ($ Z = 1 $), ground state ($ n = 1 $): \[ \lambda_H = \frac{2\pi \cdot 1 \cdot a_0}{1} = 2\pi a_0 \] Given $ \lambda_H = y $, so $ y = 2\pi a_0 $, which means $ a_0 = \frac{y}{2\pi} $.
Step 4: Calculate wavelength for He+ in the fourth orbit.
For $ He^+ $ ($ Z = 2 $), fourth orbit ($ n = 4 $): \[ \lambda_{He^+} = \frac{2\pi \cdot 4 \cdot a_0}{2} = 4\pi a_0 \]
Step 5: Express in terms of y.
Since $ a_0 = \frac{y}{2\pi} $: \[ \lambda_{He^+} = 4\pi \times \frac{y}{2\pi} = 2y \] Wait, let me recheck: $ \lambda_{He^+} = 4\pi a_0 = 4\pi \times \frac{y}{2\pi} = 2y $. But the given answer is $ y $. Re-examining: $ \lambda_H = 2\pi(1) a_0 / 1 = 2\pi a_0 = y $, so $ a_0 = y/(2\pi) $. For $ He^+ $: $ \lambda = 2\pi(4)a_0/2 = 4\pi a_0 = 4\pi \times y/(2\pi) = 2y $. The sol states answer 3 which is $ y $. Using the ratio directly: $ \lambda \propto n/Z $, ratio $ = (n_{He}/Z_{He}) / (n_H/Z_H) = (4/2)/(1/1) = 2 $, giving $ 2y $. The answer key says option 3 = $ y $, so there is a discrepancy in the problem's answer key. Following the standard derivation, the answer is $ 2y $ by applying $ \lambda \propto n/Z $.
Step 6: Final answer per standard derivation.
\[ \lambda_{He^+,\,n=4} = \frac{n}{Z}\cdot\frac{Z_H}{n_H}\cdot\lambda_H = \frac{4}{2}\cdot\frac{1}{1}\cdot y = 2y \] However, if the problem intends $ n/Z $ for $ He^+ $ = $ 4/2 = 2 $ and for H ground state = $ 1/1 = 1 $, giving $ 2y $, yet the official answer is $ y $, the answer key may have a printing error. The correct result from de Broglie quantization is: \[ \boxed{2y} \]