Question

X\(^{a+}\) and Y\(^{b+}\) are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers \(n = 1\) and \(n = 2\) of X\(^{a+}\) is \(\lambda\). The wavelength of light absorbed during the transition between the states with principal quantum numbers \(n = 2\) and \(n = 4\) of Y\(^{b+}\) is \(9\lambda\). The lowest possible value of \((a+b)\) is _______.

Hint:

Remember the Rydberg formula for hydrogen-like species. Focus on the $Z^2$ term. When relating wavelengths, setting up a ratio of the Rydberg formulas often leads to simple relationships between atomic numbers. The charges $a$ and $b$ are always one less than the atomic number $Z$.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
The problem involves two hydrogen-like species, X\(^{a+}\) and Y\(^{b+}\), undergoing electronic transitions. We are given the wavelengths of absorbed light for specific transitions and need to find the lowest possible value of \((a+b)\).

Step 2: Key Formula or Approach:
1. Rydberg formula for wavelength of absorbed light (transition):
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where \( R_H \) is the Rydberg constant, \( Z \) is the atomic number of the hydrogen-like species, and \( n_1, n_2 \) are the principal quantum numbers of the initial and final states ($n_2 > n_1$ for absorption).
2. The charge of a hydrogen-like species X\(^{a+}\) means its nuclear charge is \( Z_X \) and it has only one electron. So, \( a = Z_X - 1 \). Similarly, \( b = Z_Y - 1 \).

Step 3: Detailed Explanation:
Let the atomic number of X\(^{a+}\) be \( Z_X \) and that of Y\(^{b+}\) be \( Z_Y \).
For X\(^{a+}\):
- Transition from \( n_1 = 1 \) to \( n_2 = 2 \).
- Wavelength absorbed = \(\lambda\).
Using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H Z_X^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \frac{1}{\lambda} = R_H Z_X^2 \left( 1 - \frac{1}{4} \right) = R_H Z_X^2 \left( \frac{3}{4} \right) \]
\[ \frac{1}{\lambda} = \frac{3R_H Z_X^2}{4} \] (Equation 1)
For Y\(^{b+}\):
- Transition from \( n_1 = 2 \) to \( n_2 = 4 \).
- Wavelength absorbed = \(9\lambda\).
Using the Rydberg formula:
\[ \frac{1}{9\lambda} = R_H Z_Y^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{9\lambda} = R_H Z_Y^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R_H Z_Y^2 \left( \frac{4-1}{16} \right) = R_H Z_Y^2 \left( \frac{3}{16} \right) \]
\[ \frac{1}{9\lambda} = \frac{3R_H Z_Y^2}{16} \] (Equation 2)
Divide Equation 1 by Equation 2:
\[ \frac{1/\lambda}{1/(9\lambda)} = \frac{\frac{3R_H Z_X^2}{4}}{\frac{3R_H Z_Y^2}{16}} \]
\[ 9 = \frac{Z_X^2/4}{Z_Y^2/16} = \frac{Z_X^2}{4} \times \frac{16}{Z_Y^2} \]
\[ 9 = \frac{4Z_X^2}{Z_Y^2} \]
Rearrange to find the ratio of atomic numbers:
\[ \frac{Z_Y^2}{Z_X^2} = \frac{4}{9} \]
Taking the square root of both sides:
\[ \frac{Z_Y}{Z_X} = \frac{2}{3} \]
So, $3Z_Y = 2Z_X$.
Since $Z_X$ and $Z_Y$ must be integers (atomic numbers), the lowest possible integer values are $Z_X = 3$ and $Z_Y = 2$.
Calculate \(a\) and \(b\):
For X\(^{a+}\), $a = Z_X - 1 = 3 - 1 = 2$.
For Y\(^{b+}\), $b = Z_Y - 1 = 2 - 1 = 1$.
Calculate \(a+b\):
The lowest possible value of \((a+b) = 2 + 1 = 3\).

Step 4: Final Answer:
The lowest possible value of \((a+b)\) is 3.