Hydrogen ($^1H$), Deuterium ($^2H$), singly ionised helium ($He^+$), and doubly ionised lithium ($Li^{2+}$) al

Question

Hydrogen ($^1H$), Deuterium ($^2H$), singly ionised helium ($He^+$), and doubly ionised lithium ($Li^{2+}$) all have one electron around the nucleus. Consider an electron transition from n=2 to n=1. If the wavelengths of emitted radiations are $\lambda_1$, $\lambda_2$, $\lambda_3$, and $\lambda_4$ respectively, then approximately which one of the following is correct?

Accepted Answer

According to the Bohr model for hydrogen-like species (one-electron systems), the energy of a transition is given by $\Delta E \propto Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$. Since the wavelength $\lambda$ is inversely proportional to energy ($\Delta E = hc/\lambda$), the relationship is $\lambda \propto \frac{1}{Z^2}$ for the same transition ($n=2 	o 1$).

1.  **Hydrogen ($^1H$):** $Z=1$, so $\lambda_1 \propto 1$.
2.  **Deuterium ($^2H$):** $Z=1$ (isotope effects on Rydberg constant are negligible for this approximation), so $\lambda_2 \approx \lambda_1$.
3.  **Helium ion ($He^+$):** $Z=2$, so $\lambda_3 \propto \frac{1}{2^2} = \frac{1}{4}$. This implies $\lambda_3 = \frac{\lambda_1}{4}$ or $\lambda_1 = 4\lambda_3$.
4.  **Lithium ion ($Li^{2+}$):** $Z=3$, so $\lambda_4 \propto \frac{1}{3^2} = \frac{1}{9}$. This implies $\lambda_4 = \frac{\lambda_1}{9}$ or $\lambda_1 = 9\lambda_4$.

Combining these, we get $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

Step-by-Step Solution

Practice All PHYSICS Questions