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Q12.6(a):
(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the ((n)) = 1, 2, and 3 levels.

Solution :

Step 1: Establishing the Theoretical Framework

According to Bohr’s postulate of quantization of angular momentum, an electron in a hydrogen atom moves in a circular orbit such that its angular momentum is an integral multiple of $\frac{h}{2\pi}$. The velocity $v_n$ of an electron in the $n^{th}$ orbit is derived from the balance between the electrostatic force of attraction and the necessary centripetal force:

$\frac{mv_n^2}{r_n} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2}{r_n^2}$

By applying the quantization condition $mvr = \frac{nh}{2\pi}$, the expression for velocity in the $n^{th}$ orbit for a hydrogen atom ($Z=1$) is given by:

$v_n = \frac{e^2}{2\epsilon_0 nh}$

Substituting the universal constants:

  • Charge of electron ($e$) = $1.6 \times 10^{-19}\ C$
  • Permittivity of free space ($\epsilon_0$) = $8.854 \times 10^{-12}\ C^2N^{-1}m^{-2}$
  • Planck’s constant ($h$) = $6.626 \times 10^{-34}\ Js$
[Per Bohr's Theory of Atomic Structure]

Step 2: Calculating the Velocity Constant

To simplify, we utilize the relationship between the velocity of the first orbit ($v_1$) and the fine-structure constant ($\alpha$): $v_1 = \frac{e^2}{2\epsilon_0 h} \approx 2.188 \times 10^6\ m/s$

The velocity for any level $n$ is represented as: $v_n = \frac{v_1}{n}$

Step 3: Calculating Velocity for Specific Levels

Energy Level ($n$) Calculation Formula Calculated Speed ($m/s$)
$n = 1$ $v_1 = \frac{2.188 \times 10^6}{1}$ $2.188 \times 10^6\ m/s$
$n = 2$ $v_2 = \frac{2.188 \times 10^6}{2}$ $1.094 \times 10^6\ m/s$
$n = 3$ $v_3 = \frac{2.188 \times 10^6}{3}$ $7.293 \times 10^5\ m/s$

Step 4: Visualizing the Orbital Velocity Decay

n=1 n=2 n=3 v=2.19e6 v=1.09e6 v=0.73e6

Final Solution: The speeds of the electron in the $n=1, 2,$ and $3$ levels are $2.188 \times 10^6\ m/s$, $1.094 \times 10^6\ m/s$, and $7.293 \times 10^5\ m/s$ respectively.


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