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 λ. The wavelength of light absorbed during the transition between the states with principal quantum numbers n = 2 and n = 4 of Y^b+ is 9λ. The lowest possible value of (a+b) is _______.

Correct Answer: The lowest possible value of (a+b) is 3.

Step-by-Step Explanation:

For a hydrogen-like species with atomic number Z, the wavelength $\mapsto$ of light absorbed or emitted during a transition between energy levels with principal quantum numbers $n_1$ and $n_2$ is given by the Rydberg formula:

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where R is the Rydberg constant and Z is the nuclear charge (atomic number) of the hydrogen-like ion.

Step 1: Determine the relation for species X^a+

For X^a+, the transition occurs from $n_1=1$ to $n_2=2$, and the wavelength absorbed is $\lambda$.

Let the atomic number of X be $Z_1$. The species is X^a+, so its charge is +a, which means it has lost a electrons. Since it is a hydrogen-like species (1 electron), its nuclear charge is:

$Z_1=a+1$

Using the Rydberg formula:

$\frac{1}{\lambda }=R{Z}_1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R{Z}_1^2\left(1-\frac{1}{4}\right)=\frac{3}{4}R{Z}_1^2$

--- (Equation 1)

Step 2: Determine the relation for species Y^b+

For Y^b+, the transition occurs from $n_1=2$ to $n_2=4$, and the wavelength absorbed is 9$\lambda$.

Let the atomic number of Y be $Z_2$. Since Y^b+ is a hydrogen-like species (1 electron), its nuclear charge is:

$Z_2=b+1$

Using the Rydberg formula:

$\frac{1}{9\lambda }=R{Z}_2^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R{Z}_2^2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3}{16}R{Z}_2^2$

--- (Equation 2)

Step 3: Compare Equation 1 and Equation 2

Dividing Equation 1 by Equation 2:

$\frac{1/\lambda }{1/\left(9\lambda \right)}=\frac{\frac{3}{4}R{Z}_1^2}{\frac{3}{16}R{Z}_2^2}$

$9=4\times \frac{Z_1^2}{Z_2^2}$

Taking the square root on both sides:

$3=2\times \frac{Z_1}{Z_2}\Rightarrow \frac{Z_1}{Z_2}=\frac{3}{2}$

Step 4: Find the lowest possible values of a and b

Since $Z_1$ and $Z_2$ are atomic numbers (positive integers), the minimum possible integral values satisfying $\frac{Z_1}{Z_2}=\frac{3}{2}$ are:

$Z_1=3\text{and}{Z}_2=2$

Since $Z_1=a+1$ and $Z_2=b+1$:

$a=Z_1-1=3-1=2$

$b=Z_2-1=2-1=1$

Thus, the lowest possible value of (a+b) is:

$a+b=2+1=3$