The angular momentum of an electron in a hydrogen atom is proportional to: (Where $ r $ is the radius of the orbit of the electron)

Question

$Li^{+2} \longrightarrow Li^{+3} + e^-$ :
Energy required for this process.
Given ionisation energy for ground state of hydrogen atom is $2.17 \times 10^{-18} \text{ J}$.

Answer 1

Bohr's model of the hydrogen atom posits that an electron's angular momentum $ L $ in orbit is quantized, defined by $ L = n\hbar $. Here, $ n $ represents the principal quantum number, and $ \hbar $ denotes the reduced Planck’s constant. The radius of the $ n $-th orbit in a hydrogen atom follows $ r_n \propto n^2 $. Consequently, $ n $ can be expressed in relation to $ r $ as $ n \propto \sqrt{r} $. Substituting this into the angular momentum equation yields $ L \propto n \propto \sqrt{r} $. Therefore, the angular momentum of an electron in a hydrogen atom is directly proportional to $ \sqrt{r} $.

The angular momentum of an electron in a hydrogen atom is proportional to: (Where \( r \) is the radius of the orbit of the electron)

The Correct Option is A

Solution and Explanation

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