Question

If a hydrogen atom is its ground state it absorbs $10.2eV$ of energy. The orbital angular momentum is increased by:
A. $1.05 \times {10^{ - 34}}J/s$
B. $3.16 \times {10^{ - 34}}J/s$
C. $2.11 \times {10^{ - 34}}J/s$
D. $4.22 \times {10^{ - 34}}J/s$

Answer 1

Hint:Here the angular momentum of a hydrogen atom in its second orbit. The angular momentum of an electron, according to Bohr's second postulate, is a multiple of $2h$ which is integral. Using this, we can calculate the second orbit's angular momentum in terms of $L$. The third orbit's angular momentum in terms of $L$ can then be determined using the same method.

Formula used:
The formula of angular momentum in ${n^{th}}$ orbit is given by:
${L_n} = n\dfrac{h}{{2\pi }}$
Here, $h$ is the Planck constant,

Complete step by step solution:
In order to know that each orbit in which the electron circulates around the nucleus is discrete and is characterized by a particular main quantum number$(n)$ as according to Bohr's atomic model. And, Bohr’s theory suggests that the angular momentum of an electron is given by:
${L_n} = n\dfrac{h}{{2\pi }}$

In the question, we have given that an electron from the ground state- observed $10.2eV$ of energy jumps from the ground state $n = 1$ to the first excited state $n = 2$. But the hydrogen atom is described by Bohr's model as:
$L = \dfrac{{nh}}{{2\pi }}$
Now, substitute the value $n = 1$ in the above formula, then we have:
${L_1} = \dfrac{{(1)h}}{{2\pi }} \\
\Rightarrow \dfrac{h}{{2\pi }}\,\,\,\,....(i)$

Similarly, substitute the value $n = 2$, then:
${L_2} = \dfrac{{(2)h}}{{2\pi }} \\
\Rightarrow \dfrac{h}{\pi }\,\,\,\,...(ii)$
Subtracting the obtained equations $(i)$ from $(ii)$to determine the change in momentum, we have:
$\Delta L = {L_2} - {L_1} \\$
$\Rightarrow \Delta L = \dfrac{h}{\pi } - \dfrac{h}{{2\pi }} \\$
$\Rightarrow \Delta L = \dfrac{h}{{2\pi }} \\$
As we know that the value of Planck constant $h = 6.63 \times {10^{ - 34}}\,J/s$, so we can substitute the value of $h$in the above equation, we obtain:
$\Delta L = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{2 \times 3.14}} \\$
$\Rightarrow \Delta L = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{6.28}} \\$
$\therefore \Delta L = 1.05 \times {10^{ - 34}}J/s \\$
Thus, the correct option is A.

Note: It should be noted that Bohr's atomic model states that electrons orbiting a nucleus have quantized angular momentum and can only move in orbits where their angular momentum is a multiple of $\dfrac{h}{2}$. The De Broglie equation is used to describe this theory on the quantization of an electron's angular momentum.