If the ionization energy of a hydrogen atom in its ground state is denoted as , the energy required — Atomic Structure Chemistry Question
Question
If the ionization energy of a hydrogen atom in its ground state is denoted as $E_0$, the energy required for its electron to jump from the $2^{nd}$ orbit to the $3^{rd}$ orbit can be expressed as $\frac{y}{36} E_0$. Find the integer value of $y$.
Answer: 5
💡 Solution & Explanation
The ionization energy $E_0 = 13.6 \text{ eV}$. The energy required to jump from $n=2$ to $n=3$ is $\Delta E = E_0 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = E_0 \times (\frac{1}{2^2} - \frac{1}{3^2}) = E_0 \times (\frac{1}{4} - \frac{1}{9}) = E_0 \times \frac{9-4}{36} = \frac{5}{36} E_0$. Thus, $y = 5$.
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