Question

The energy of an electron in an excited hydrogen atom is −3.4 eV. Then, according to Bohr's Theory, the angular momentum of this electron, in J s, is
A. \[2.11 \times {10^{ - 34}}\]
B. \[3 \times {10^{ - 34}}\]
C. \[1.055 \times {10^{ - 34}}\]
D. \[0.5 \times {10^{ - 34}}\]

Answer 1

Hint: An electron's angular momentum in the orbit is quantized. Find out the orbit number in which an electron is situated by putting the value of given energy in the formula \[E = - \dfrac{{13.6}}{{{n^2}}}eV\]. Then when we get value of $n$ and can easily find out the angular momentum by the formula \[mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}\].

Formula used:
Energy of the electron in nth orbit is,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
The formula of angular momentum ($L$) is,
 \[L= {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Here, $h$ is the planck’s constant and $n$ is the orbit number.

Complete step by step solution:
Three postulates form the foundation of Bohr's model of the hydrogen atom:
(1) Around the nucleus, an electron moves in a circular orbit.;
(2) The quantization of an electron's orbital angular momentum; and
(3) The electrons in an atom move from a lower energy level to a higher energy level by gaining the necessary energy, and an electron moves from a higher energy level to a lower energy level by losing energy.

Let's discuss the second postulates in more detail. The second premise discusses stable orbits. Bohr claimed that electrons can rotate in specific discrete (discontinuous), non-radiating orbits known as stationary (permitted) orbits, for which the angular momentum of the rotating electron (L) is an integral multiple of $\dfrac{h}{{2\pi }}$.

As a result, the circling electron's angular momentum is an integral multiple of integer. An electron's angular momentum is equal to \[mvr\]. For every orbit that is stationary:
\[mvr{\text{ }} = {\text{ }}\dfrac{{nh}}{{2\pi }}\]
Where, n = a main quantum number with a positive integral value between 1 and n with a range of 1, 2, and 3. Planck's constant, h, is equal to \[6.63{\text{ }} \times {\text{ }}{10^{ - 34}}{\text{ }}Js\].

The energy in any orbit for a hydrogen atom can be defined as,
\[E = - \dfrac{{13.6}}{{{n^2}}}eV\]
Here energy is given -3.4 eV. So, we can write
\[E = \dfrac{{ - 13.6}}{{{n^2}}}eV \\
\Rightarrow - 3.4 = - \dfrac{{13.6}}{{{n^2}}} \\ \]
$\Rightarrow {n^2} = \dfrac{{ - 13.6}}{{ - 3.4}} = 4 \\
\Rightarrow n = 2 \\ $
Now we have to find out the angular momentum of Bohr’s second’s orbit. So, we already see a formula for finding angular momentum. If we put the given values then we get.
${\text{Angular momentum = }}\dfrac{{{\text{nh}}}}{{2\pi }} \\
\Rightarrow \text{Angular momentum} = \dfrac{{2 \times 6.63 \times {{10}^{ - 34}}}}{{2 \times 3.14}} \\
\therefore \text{Angular momentum}= 2 \times {10^{ - 34}}kg - {m^2}/\sec $

Hence, option A is the correct option.

Notes The primary distinction between angular momentum and linear momentum is that linear momentum is a property of moving objects relative to a reference point (i.e., any object changing its position with respect to the reference point), angular momentum is a property of moving objects relative to a reference point that also involves changing their position's direction (i.e. they are not moving in a straight line).