For a single ionized helium atom ,the longest wavelength in ground state absorb will be?

$\left(a\right) 1216 \times {10}^{-} 10 m$ $\left(b\right) 912 \times {10}^{-} 10 m$ $\left(c\right) 304 \times {10}^{-} 10 m$ $\left(d\right) 604 \times {10}^{-} 10 m$

Jun 29, 2018

(c) $\lambda = 304 \times {10}^{- 10} \textcolor{w h i t e}{l} m$

Explanation:

A helium atom $\text{He}$ contains two electrons. Removing one of the two electrons produces a ${\text{He}}^{+}$ ion. The ion would be left with one single electron and is isoelectric to a hydrogen atom.

That electron would experience no electrostatic repulsion from other atomic electrons (since there were none.) As a result, the nuclei's attraction on that electron would be the only interaction influencing the amount of energy necessary to promote the electron by the minimum extent and hence the longest absorption wavelength possible.

The Rydberg Formula enables the calculation of this absorption wavelength (along with that of any other possible transition) without knowledge of the radius of the electron cloud as long as the following values are available

• The Rydberg Number $\text{R} \approx 1.097 \times {10}^{7} \textcolor{w h i t e}{l} {m}^{- 1}$
• The atomic number; $\text{Z} = 2$ for helium
• The principal quantum number of orbitals the electron occupied before and after the transition of interest. Radiations of long wavelength correspond to those of low energy. The electron currently sits at $n = 1$, a transition to $n = 2$ allows for the minimum energy change. ${n}_{1} = 2$ and ${n}_{2} = 1$. ${n}_{1} = 2$ and ${n}_{2} = 1$.

$\frac{1}{\lambda} = R \cdot {Z}^{2} \cdot \left(\frac{1}{{n}_{1}^{2}} - \frac{1}{{n}_{2}^{2}}\right)$

lambda = 1/ (R*Z^2*(1"/"n_1^2-1"/"n_2^2)) = 3.04 xx 10^8 color(white)(l) "m" = 304 color(white)(l) "n"m