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Chapter 27: Problem 50

By directly substituting the values of the fundamental constants, show that the ground state energy for hydrogen in the Bohr model $E_{1}=-m_{\mathrm{e}} k^{2} e^{4} /\left(2 \hbar^{2}\right)$ has the numerical value -13.6 eV.

Short Answer

Expert verified
Answer: The ground state energy for hydrogen in the Bohr model is -13.6 eV.

Step by step solution

01

Write down the values of the fundamental constants

Following are the given constants: - Mass of an electron: \(m_{\mathrm{e}} = 9.109 \times 10^{-31} \, \mathrm{kg}\) - Coulomb constant: \(k = 8.987 \times 10^{9} \, \mathrm{N\cdot m^{2} \cdot C^{-2}}\) - Elementary charge: \(e = 1.602 \times 10^{-19} \, \mathrm{C}\) - Reduced Planck constant: \(\hbar = 1.055 \times 10^{-34} \, \mathrm{J\cdot s}\)
02

Substitute the values into the ground state energy formula

Plug in the values of the fundamental constants into the formula: \(E_{1}=-m_{\mathrm{e}} k^{2} e^{4} /\left(2 \hbar^{2}\right) = -9.109 \times 10^{-31} \left(8.987\times 10^{9}\right)^2 \times \left(1.602 \times 10^{-19}\right)^4 / \left(2 \times \left(1.055 \times 10^{-34}\right)^2\right)\).
03

Calculate the energy value in joules

Now, perform the calculation: \(E_{1} = -9.109 \times 10^{-31} \times 8.083 \times 10^{20} \times 6.571 \times 10^{-76} / (2 \times 1.114 \times 10^{-68}) \approx -2.178 \times 10^{-18} \, \mathrm{J}\).
04

Convert the energy value from joules to electronvolts

To convert the energy value from joules to electronvolts, divide by the elementary charge value: \(E_{1} = -2.178 \times 10^{-18} \, \mathrm{J} / \left(1.602 \times 10^{-19} \, \mathrm{C}\right) \approx -13.6 \, \mathrm{eV}\). Therefore, the ground state energy for hydrogen in the Bohr model is indeed -13.6 eV.

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Most popular questions from this chapter

In gamma-ray astronomy, the existence of positrons \(\left(\mathrm{e}^{+}\right)\) can be inferred by a characteristic gamma ray that is emitted when a positron and an electron (e \(^{-}\) ) annihilate. For simplicity, assume that the electron and positron are at rest with respect to an Earth observer when they annihilate and that nothing else is in the vicinity. (a) Consider the reactions $\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow \gamma$, where the annihilation of the two particles at rest produces one photon, and \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma,\) where the annihilation produces two photons. Explain why the first reaction does not occur, but the second does. (b) Suppose the reaction \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma\) occurs and one of the photons travels toward Earth. What is the energy of the photon?
How much energy is required to remove an electron from a hydrogen atom in the \(n=4\) state?
An x-ray photon with wavelength 6.00 pm collides with a free electron initially at rest. What is the maximum possible kinetic energy acquired by the electron?
An incident photon of wavelength \(0.0100 \mathrm{nm}\) is Compton scattered; the scattered photon has a wavelength of \(0.0124 \mathrm{nm} .\) What is the change in kinetic energy of the electron that scattered the photon?
The Bohr theory of the hydrogen atom ignores gravitational forces between the electron and the proton. Make a calculation to justify this omission. [Hint: Find the ratio of the gravitational and electrostatic forces acting on the electron due to the proton.]
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