Atomic Structure: Practice Problem & Solution
The energy of hydrogen atom in $n^{th}$ orbit is $E_n$ then the energy in $n^{th}$ orbit of singly ionised helium atom will be: (2001)
Solution Explained:
To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:
Energy of an electron in a hydrogen-like atom is $E \propto Z^2$. For hydrogen $Z=1$, $E = E_n$. For singly ionised helium ($He^+$), $Z=2$. Therefore, the energy in the same orbit is $2^2 E_n = 4 E_n$.
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