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Chapter 16: Problem 79

The Bohr model of the hydrogen atom proposes that the electron orbits around the proton in a circle of radius \(5.3 \times 10^{-11} \mathrm{m} .\) The electric force is responsible for the radial acceleration of the electron. What is the speed of the electron in this model?

Short Answer

Expert verified
Answer: The speed of the electron in the hydrogen atom using the Bohr model is approximately \(2.18 \times 10^6 \;\text{m/s}\).

Step by step solution

01

Write down the given information

We are given the radius of the electron's orbit in the hydrogen atom: \(r = 5.3 \times 10^{-11} \mathrm{m}\).
02

Equate the centripetal force and electric force

The centripetal force on the electron is provided by the electric force between the electron and proton. Thus, we can equate the two forces: $$m \times a_c = \frac{k \times e^2}{r^2}$$, where \(a_c = \frac{v^2}{r}\) is the centripetal acceleration, and \(v\) is the speed we are looking for.
03

Substitute the centripetal acceleration

Replace \(a_c\) with \(\frac{v^2}{r}\) in the above equation: $$m \times \frac{v^2}{r} = \frac{k \times e^2}{r^2}$$.
04

Rearrange the equation to solve for \(v\)

We need to isolate \(v\) so we can find its value: $$v^2 = \frac{k \times e^2 \times r}{m}$$.
05

Substitute the known values and constants

Apply the known values and constants to the equation: $$v^2 = \frac{(8.99\times10^9\;\mathrm{N\cdot m^2/C^2})\times (1.6\times10^{-19}\;\mathrm{C})^2 \times (5.3\times10^{-11}\;\mathrm{m})}{9.11\times10^{-31}\;\mathrm{kg}}$$.
06

Calculate the value of \(v\)

Perform the calculation to find the value of \(v\): $$v=\sqrt{\frac{(8.99\times10^9\;\mathrm{N\cdot m^2/C^2})\times (1.6\times10^{-19}\;\mathrm{C})^2 \times (5.3\times10^{-11}\;\mathrm{m})}{9.11\times10^{-31}\;\mathrm{kg}}}$$, which yields \(v \approx 2.18 \times 10^6 \;\text{m/s}\). Solution: The speed of the electron in the hydrogen atom using the Bohr model is approximately \(2.18 \times 10^6 \;\text{m/s}\).

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