Question 503: Atoms & Nuclei

Question

The energy of an electron in the excited hydrogen atom is $$-3.4 \mathrm{~eV}$$. Then according to Bohr's theory, the angular momentum of the electron in that excited state is ($$\mathrm{h}=$$ Plank's constant)

Accepted Answer

According to Bohr's theory, the energy levels of an electron in a hydrogen atom are given by:
$$E_n = -\frac{13.6 \, 	ext{eV}}{n^2}$$
Given that the energy of the electron in the excited state is $$-3.4 \, 	ext{eV}$$, we can set up the following equation to find the principal quantum number $$n$$:
$$-3.4 = -\frac{13.6}{n^2}$$
Solving for $$n$$, we get:
$$n^2 = \frac{13.6}{3.4} = 4$$
So, $$n = 2$$. Now, according to Bohr's theory, the angular momentum $$L$$ of the electron in the $$n$$th excited state is given by:
$$L = n \frac{h}{2 \pi}$$
By substituting the value of $$n = 2$$, we get:
$$L = 2 \frac{h}{2 \pi} = \frac{h}{\pi}$$
Therefore, the angular momentum of the electron in that excited state is:
Option C
$$\frac{\mathrm{h}}{\pi}$$

Answer

$$\frac{2 \pi}{h}$$

Answer

$$\frac{\mathrm{nh}}{2 \pi}$$

Answer

$$\frac{\mathrm{h}}{\pi}$$

Answer

$$\frac{3 \mathrm{~h}}{2 \pi}$$

Atoms & Nuclei

The energy of an electron in the excited hydrogen atom is $-3.4 \mathrm{~eV}. Then according to Bohr's theory, the angular momentum of the electron in that excited state is (\mathrm{h}=$ Plank's constant)

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