Question

Express the speed of the electron in the Bohr model in terms of the fundamental constants (${\mathrm{m}}_{\mathrm{e}}$, e, h, ${\mathrm{\epsilon }}_0$), the nuclear charge Z, and the quantum number n. Evaluate the speed of an electron in the ground states of${\text{He}}^{+}$ion and${\text{U}}^{91+}$. Compare these speeds with the speed of light c. As the speed of an object approaches the speed of light, relativistic effects become important. In which kinds of atoms do you expect relativistic effects to be greatest?

Short answer

The speed of the electron in the ground state of ${\text{He}}^{+}$ions is $4.38\times {10}^6\text{m}$and ${\text{U}}^{91+}$is $1.99\times {10}^8\text{m}$and the relativistic effect is more pronounced for the heavier atom as ${\text{U}}^{91+}$.

Step-by-step solution

The Concept of Bohr Atomic Theory

In Bohr’s atomic theory for hydrogen-like species, the velocity of the electron in $n^{\text{th}}$orbit is as follows:

$v_n=\frac{Z{e}^2}{2{\epsilon }_{0}nh}$

Where,

- Z is the atomic number.

- e is a charge on an electron.

-${\epsilon }_0$is the permittivity of space.

- n is the principle quantum number.

- h is Plank's constant.

Calculating the speed of an electron in the ground state of helium ions

For helium,

• The value ofZis2.
• The value of is $1.6\times {10}^{-19}\text{C}$.
• The value of is $8.85\times {10}^{-12}{\text{m}}^{-3}·{\text{kg}}^{-1}·{\text{s}}^4·{\text{A}}^2$.
• The value of n is 1.
• The value of h is $6.626\times {10}^{-34}\text{J}·\text{s}$.

Substitute these values in the above equation.

$$\begin{gathered} v_n=\frac{Z{e}^2}{2{\epsilon }_{0}nh} \\ {v}_n=\frac{\left(2\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{2\left(8.85\times {10}^{-12}{\text{m}}^{-3}·{\text{kg}}^{-1}·{\text{s}}^4·{\text{A}}^2\right)\left(1\right)\left(6.626\times {10}^{-34}\text{J}-\text{s}\right)} \\ {v}_n=4.38\times {10}^6\text{m} \end{gathered}$$

Calculating the Speed of an Electron in the Ground State of Uranium Ions

For uranium,

• The value of Z is 92 .
• The value of e is$1.6\times {10}^{-19}\text{C}$
• The value of ${\epsilon }_0$ is$8.85\times {10}^{-12}{\text{m}}^{-3}·{\text{kg}}^{-1}·{\text{s}}^4·{\text{A}}^2$.
• The value of n is 1.
• The value of h is$6.626\times {10}^{-34}\text{J}·\text{s}$

Substitute these values in the equation above equation.

$$\begin{gathered} v_n=\frac{Z{e}^2}{2e_{0}nh} \\ {v}_n=\frac{\left(92\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{2\left(8.85\times {10}^{-12}{\text{m}}^{-3}·{\text{kg}}^{-1}·{\text{s}}^4·{\text{A}}^2\right)\left(1\right)\left(6.626\times {10}^{-34}\text{J}·\text{s}\right)} \\ {v}_n=1.99\times {10}^8 \end{gathered}$$

Since the velocity of electron increases with an increase in mass of the atom and thus, the relativistic effect is more pronounced for the heavier atom as ${\text{U}}^{91+}$.

Thus,the speed of an electron in the ground state of${\mathrm{He}}^{+}$ ions is$4.38\times {10}^6\text{m}$and${\text{U}}^{91+}$is $1.99\times {10}^8\text{m}$. The relativistic effect is more pronounced for the heavier atom as${\text{U}}^{91+}$ .