Question

Which of the following transitions of $\text{He}^+$ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom?

• A. $n=4$ to $n=2$
• B. $n=6$ to $n=5$
• C. $n=6$ to $n=3$
• D. None of these

Answer 1

Answer: (A)

A

Answer 2

The energy of a spectral line transition in a hydrogen-like atom/ion with atomic number $Z$ is given by:

$\Delta E = R_H \cdot Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

where $R_H$ is the Rydberg constant, $n_i$ is the initial principal quantum number, and $n_f$ is the final principal quantum number ($n_i > n_f$ for emission).

The wavelength $\lambda$ of the emitted photon is related to the energy difference by $\Delta E = \frac{hc}{\lambda}$.

Thus, $\frac{1}{\lambda} = \frac{\Delta E}{hc} = \frac{R_H}{hc} \cdot Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$.

Let $R = \frac{R_H}{hc}$ be the Rydberg constant in terms of inverse wavelength.

So, $\frac{1}{\lambda} = R \cdot Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$.

For a hydrogen atom, $Z=1$. The reciprocal wavelength is $\frac{1}{\lambda_H} = R \cdot 1^2 \left(\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}\right) = R \left(\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}\right)$.

For a $\text{He}^+$ ion, $Z=2$. The reciprocal wavelength is $\frac{1}{\lambda_{He^+}} = R \cdot 2^2 \left(\frac{1}{n_{f,He^+}^2} - \frac{1}{n_{i,He^+}^2}\right) = 4R \left(\frac{1}{n_{f,He^+}^2} - \frac{1}{n_{i,He^+}^2}\right)$.

We want to find a transition in $\text{He}^+$ that gives the same wavelength as a spectral line in hydrogen.

So, $\lambda_{He^+} = \lambda_H$, which implies $\frac{1}{\lambda_{He^+}} = \frac{1}{\lambda_H}$.

$4R \left(\frac{1}{n_{f,He^+}^2} - \frac{1}{n_{i,He^+}^2}\right) = R \left(\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}\right)$

$4 \left(\frac{1}{n_{f,He^+}^2} - \frac{1}{n_{i,He^+}^2}\right) = \left(\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}\right)$

We need to check each option for the transition in $\text{He}^+$ and see if the resulting value on the left side can be represented as $\left(\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}\right)$ for some integers $n_{i,H} > n_{f,H} \ge 1$.

A) $n=4$ to $n=2$ in $\text{He}^+$.

$n_{i,He^+} = 4$, $n_{f,He^+} = 2$.

LHS = $4 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4 \left(\frac{1}{4} - \frac{1}{16}\right) = 4 \left(\frac{4-1}{16}\right) = 4 \times \frac{3}{16} = \frac{3}{4}$.

We need to check if $\frac{3}{4} = \frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}$ for integers $n_{i,H} > n_{f,H} \ge 1$.

If we take $n_{f,H} = 1$, we get $\frac{1}{1^2} - \frac{1}{n_{i,H}^2} = 1 - \frac{1}{n_{i,H}^2} = \frac{3}{4}$.

$1 - \frac{3}{4} = \frac{1}{n_{i,H}^2} \implies \frac{1}{4} = \frac{1}{n_{i,H}^2} \implies n_{i,H}^2 = 4$.

Since $n_{i,H}$ must be a positive integer, $n_{i,H} = 2$.

We have $n_{f,H} = 1$ and $n_{i,H} = 2$, which is a valid transition in hydrogen (from $n=2$ to $n=1$).

Thus, the transition $n=4$ to $n=2$ in $\text{He}^+$ has the same wavelength as the transition $n=2$ to $n=1$ in hydrogen.

B) $n=6$ to $n=5$ in $\text{He}^+$.

$n_{i,He^+} = 6$, $n_{f,He^+} = 5$.

LHS = $4 \left(\frac{1}{5^2} - \frac{1}{6^2}\right) = 4 \left(\frac{1}{25} - \frac{1}{36}\right) = 4 \left(\frac{36-25}{900}\right) = 4 \times \frac{11}{900} = \frac{11}{225}$.

We need $\frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2} = \frac{11}{225}$.

$\frac{n_{i,H}^2 - n_{f,H}^2}{n_{f,H}^2 n_{i,H}^2} = \frac{11}{225}$.

$225(n_{i,H}^2 - n_{f,H}^2) = 11 n_{f,H}^2 n_{i,H}^2$.

Since 11 is prime, either $n_{f,H}^2$ or $n_{i,H}^2$ must be a multiple of 11. Also, the denominator $n_{f,H}^2 n_{i,H}^2$ must be a multiple of 225.

Let's test possible integer values for $n_{f,H}$ and $n_{i,H}$.

For $n_{f,H}=1$, $1 - 1/n_{i,H}^2 = 11/225 \implies 1/n_{i,H}^2 = 1 - 11/225 = 214/225 \implies n_{i,H}^2 = 225/214$, not an integer.

For $n_{f,H}=2$, $1/4 - 1/n_{i,H}^2 = 11/225 \implies 1/n_{i,H}^2 = 1/4 - 11/225 = (225-44)/900 = 181/900 \implies n_{i,H}^2 = 900/181$, not an integer.

For $n_{f,H}=3$, $1/9 - 1/n_{i,H}^2 = 11/225 \implies 1/n_{i,H}^2 = 1/9 - 11/225 = (25-11)/225 = 14/225 \implies n_{i,H}^2 = 225/14$, not an integer.

For $n_{f,H}=4$, $1/16 - 1/n_{i,H}^2 = 11/225 \implies 1/n_{i,H}^2 = 1/16 - 11/225 = (225-176)/3600 = 49/3600 \implies n_{i,H}^2 = 3600/49 = (60/7)^2$, not an integer.

It can be shown that there are no integer solutions for $n_{f,H}, n_{i,H}$.

C) $n=6$ to $n=3$ in $\text{He}^+$.

$n_{i,He^+} = 6$, $n_{f,He^+} = 3$.

LHS = $4 \left(\frac{1}{3^2} - \frac{1}{6^2}\right) = 4 \left(\frac{1}{9} - \frac{1}{36}\right) = 4 \left(\frac{4-1}{36}\right) = 4 \times \frac{3}{36} = 4 \times \frac{1}{12} = \frac{1}{3}$.

We need $\frac{1}{3} = \frac{1}{n_{f,H}^2} - \frac{1}{n_{i,H}^2}$.

For $n_{f,H}=1$, $1 - 1/n_{i,H}^2 = 1/3 \implies 1/n_{i,H}^2 = 1 - 1/3 = 2/3 \implies n_{i,H}^2 = 3/2$, not an integer.

For $n_{f,H}=2$, $1/4 - 1/n_{i,H}^2 = 1/3 \implies 1/n_{i,H}^2 = 1/4 - 1/3 = (3-4)/12 = -1/12$. This is negative, but $1/n_{i,H}^2$ must be positive. This requires $1/n_{f,H}^2 > 1/3$, so $n_{f,H}^2 < 3$, meaning $n_{f,H}=1$. As shown, $n_{f,H}=1$ does not yield an integer $n_{i,H}$.

Thus, there are no integer solutions for $n_{f,H}, n_{i,H}$.

Only option A results in a wavelength that matches a spectral line in hydrogen.

The final answer is $\boxed{A}$.

Explanation: The wavelength of a spectral line is determined by the energy difference between the initial and final energy levels, which is given by the Rydberg formula scaled by $Z^2$. For $\text{He}^+$ ($Z=2$), the energy difference is 4 times that for hydrogen ($Z=1$) for the same transition numbers. To get the same wavelength (and thus the same energy difference), the term $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ for $\text{He}^+$ must be $1/4$ times the term for hydrogen. For option A, the transition in $\text{He}^+$ is from $n=4$ to $n=2$. The term is $\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16}$. The effective term for wavelength calculation is $4 \times \frac{3}{16} = \frac{3}{4}$. This value $\frac{3}{4}$ can be represented as $\left(\frac{1}{1^2} - \frac{1}{2^2}\right)$, which corresponds to the transition from $n=2$ to $n=1$ in hydrogen.