Question 134: Atoms & Nuclei

Question

The energy of an electron in an orbit of the Bohr's atom is $-0.04E_0$ eV where $E_0$ is the ground state energy. If $L$ is the angular momentum of the electron in this orbit and $h$ is the Planck's constant, then
$ \frac{2\pi L}{h} $ is ________ :

Accepted Answer

In the Bohr model of the atom, the energy of an electron in the $n^{	ext {th }}$ orbit $\left(E_night)$ is related to the ground state energy ( $\mathrm{E}_0$ ) by the formula :
$$ \mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{E}_0}{\mathrm{n}^2} $$
It is given that the energy in the current orbit is $-0.04 \mathrm{E}_0 \mathrm{eV}$.
$$ -0.04 \mathrm{E}_0=-\frac{\mathrm{E}_0}{\mathrm{n}^2} $$
$$\Rightarrow $$ $0.04=\frac{1}{\mathrm{n}^2}$
$$\Rightarrow $$ $\mathrm{n}^2=\frac{100}{4}=25$
$$\Rightarrow $$ $$ \mathrm{n}=\sqrt{25}=5 $$
So, the electron is in the $5^{	ext {th }}$ orbit of the Bohr's atom.
Bohr's second postulate states that the angular momentum of an electron is quantized and is an integral multiple of $\mathrm{h} / 2 \pi$ :
$$ \mathrm{L}=\frac{\mathrm{nh}}{2 \pi} $$
Where n is the principal quantum number (orbit number).
$$ \frac{2 \pi L}{h}=n=5 $$
Therefore, the value of $\frac{2 \pi L}{h}$ is 5 .
Hence, the correct option is (C).

Answer

6

Answer

2

Answer

5

Answer

4

Atoms & Nuclei

The energy of an electron in an orbit of the Bohr's atom is -0.04E_0 eV where E_0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then \frac{2\pi L}{h} is ________ :

Choose the correct answer: