Question

The ratio of the speed of the electron in the ground state of hydrogen to the speed of light in vacuum is

Answer 1

Answer on Question #41321 - Physics - Molecular Physics

Question.

The ratio of the speed of the electron in the ground state of hydrogen to the speed of light in vacuum is

$$\frac {v _ {e}}{c} = ?$$

Solution.

The speed of light in vacuum:

$$c = 3 \cdot 1 0 ^ {8} \frac {m}{s}$$

The speed of the electron in the ground state of Hydrogen will be found using Bohr's postulates.

Newton's second law of motion for an electron in a circular orbit in the Coulomb force:

$$\frac {m v ^ {2}}{r} = \frac {k e ^ {2}}{r ^ {2}} \rightarrow v = \sqrt {\frac {k e ^ {2}}{m r}}$$

$m = 9.1 \cdot 10^{-31} \, \text{kg}$ is the mass of electron;

$v = v_{e}$ is the velocity of electron;

$k = 9 \cdot 10^{9} \frac{N \cdot m^{2}}{C^{2}}$ is a Coulomb's constant;

$e = 1.6 \cdot 10^{-19} \, \text{C}$ is an electron charge;

$r$ is the radius of an orbit (for ground state);

We need to find the radius $r$ and then we find the velocity $v$ .

Use Bohr quantum rule: The angular momentum $L = mvr$ is an integer multiple of $\hbar$ :

$$m v r = n \hbar$$ $$\hbar = 1. 0 5 \cdot 1 0 ^ {- 3 4} J \cdot s$$

Obtain from these two formulas:

$$r _ {n} = \frac {n ^ {2} \hbar^ {2}}{k m e ^ {2}}$$ $$v _ {n} = \frac {k e ^ {2}}{n \hbar}$$

$n = 1$ is a ground state of Hydrogen, principal orbit.

So,

$$r _ {1} = \frac {\hbar^ {2}}{k m e ^ {2}} = 0. 5 3 \cdot 1 0 ^ {- 1 1} m$$ $$v _ {1} = v _ {e} = \frac {k e ^ {2}}{\hbar} = 2. 2 \cdot 1 0 ^ {6} \frac {m}{s}$$

Thus,

$$\frac {v _ {e}}{c} = \frac {2 . 2 \cdot 1 0 ^ {6}}{3 \cdot 1 0 ^ {8}} = 0. 7 3 \cdot 1 0 ^ {- 2} = 0. 7 3 \%$$

Answer.

$$\frac {v _ {e}}{c} = 0. 7 3 \cdot 1 0 ^ {- 2} = 0. 7 3 \%$$

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