The distance between 4th and 3rd Bohr orbits of $He^{+}$ is:
A$2.645 \times 10^{-10} \, m$
B$1.322 \times 10^{-10} \, m$
C$1.851 \times 10^{-10} \, m$
D$4.761 \times 10^{-10} \, m$
Answer & Solution
Correct answer: C. $1.851 \times 10^{-10} \, m$
For a hydrogen-like species, $r_n = 0.529\dfrac{n^2}{Z}\,\text{\AA}$. For $He^+$, $Z=2$. So $r_4-r_3=0.529\left(\dfrac{16}{2}-\dfrac{9}{2}\right)\text{\AA}=0.529\times 3.5\text{\AA}=1.851\times 10^{-10}\,\text{m}$.
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