Chapter 21: Q73P (page 629)

In an early model of the hydrogen atom (the Bohr model), the electron orbits the proton in uniformly circular motion.The radius of the circle is restricted (quantized) to certain values given by ${r=n}^{2} a_{0},forn=1,2,3,$ . . . ,where $a_{0} =52.92pm$ .What is the speed of the electron if it orbits in
(a) the smallest allowed orbit and (b) the second smallest orbit? (c)
If the electron moves to larger orbits, does its speed increase, decrease,
or stay the same?

Short Answer

Expert verified

a) The speed of the smallest allowed orbit is $v_{1} = 2.19 \times 10^{6} \frac{m}{s}$ .

b) The speed of the second smallest orbit $v_{2} = 1.09 \times 10^{6} \frac{m}{s}$ .

c)Speed of the electron will decrease if it moves to larger orbits

Step by step solution

(a) Calculate the speed of the electron if it orbits in the smallest allowed orbit

The Coulomb force between the electron and the proton provides the

centripetal force that keeps the electron in circular orbit about the proton:

$\frac{k {| e |}^{2}}{r^{2}} = \frac{m_{e} v^{2}}{r}$

The smallest orbital radius is $r_{1} = a_{0} = 52.9 \times 10^{- 12} m$ . The corresponding speed of the electron is obtained as follows:

$\begin{matrix} v_{1} = \sqrt{\frac{k {| e |}^{2}}{m_{e} r_{1}}} \\ = \sqrt{\frac{k {| e |}^{2}}{m_{e} a_{0}}} \end{matrix}$

Substitute the values and solve as:

$\begin{matrix} v_{1} = \sqrt{\frac{(9 \times 10^{9} \frac{{Nm}^{2}}{C^{2}}) {(1.60 \times 10^{- 19} C)}^{2}}{(9.11 \times 10^{- 31} kg) (52.9 \times 10^{- 12} m)}} \\ = 2.19 \times 10^{6} \frac{m}{s} \end{matrix}$

(b) Calculate the speed of the electron if it orbits in the second smallest orbit

The radius of the second smallest orbit is $r_{2} = {(2)}^{2} a_{0} = 4 a_{0}$ .

Thus, the speed of the electron is as follows:

$\begin{matrix} v_{2} = \sqrt{\frac{k {| e |}^{2}}{m_{e} r_{2}}} \\ = \sqrt{\frac{k {| e |}^{2}}{m_{e} (4 a_{0})}} \end{matrix}$

Substitute the values and solve as:

$\begin{matrix} v_{2} = \frac{1}{2} v_{1} \\ = \frac{1}{2} (2.19 \times 10^{6} \frac{m}{s}) \\ = 1.09 \times 10^{6} \frac{m}{s} \end{matrix}$

(c) Find out if the electron’s speed increases or decreases or stays the sameif the electron moves to larger orbits

Since the speed is inversely proportional to $r^{\frac{1}{2}}$ , the speed of the electron will decrease if it moves to larger orbits

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Short Answer

Step by step solution

(a) Calculate the speed of the electron if it orbits in the smallest allowed orbit

(b) Calculate the speed of the electron if it orbits in the second smallest orbit

(c) Find out if the electron’s speed increases or decreases or stays the sameif the electron moves to larger orbits

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