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Chapter 27: Problem 72

What is the ground state energy, according to Bohr theory, for (a) \(\mathrm{He}^{+},\) (b) \(\mathrm{Li}^{2+},\) (c) deuterium (an isotope of hydrogen whose nucleus contains a neutron as well as a proton)?

Short Answer

Expert verified
Based on Bohr's theory, the ground state energy for \({He}^{+}\) is approximately -54.4 eV, for \({Li}^{2+}\) is approximately -122.4 eV, and for deuterium is approximately -13.6 eV.

Step by step solution

01

Calculating ground state energy of \({He}^{+}\)#

For \({He}^{+}\), its atomic number \(Z = 2\). Since we are interested in the ground state energy, set \(n = 1\): \(E_{1} = -\dfrac{(2)^2 e^4 m_e}{2(4\pi\epsilon_o)^2\hbar^2 (1)^2}\) Now, calculate the ground state energy: \(E_{1} = -\dfrac{(2)^2 (1.602 \times 10^{-19})^4 (9.11 \times 10^{-31})}{2(4\pi(8.854 \times 10^{-12})^2(1.055 \times 10^{-34})^2}\) \(E_{1} \approx -54.4\) eV So, the ground state energy of \({He}^{+}\) is approximately -54.4 eV. #b) Ground state energy of \({Li}^{2+}\)#
02

Calculating ground state energy of \({Li}^{2+}\)#

For \({Li}^{2+}\), its atomic number \(Z = 3\). Since we are interested in the ground state energy, set \(n = 1\): \(E_{1} = -\dfrac{(3)^2 e^4 m_e}{2(4\pi\epsilon_o)^2\hbar^2 (1)^2}\) Now, calculate the ground state energy: \(E_{1} = -\dfrac{(3)^2 (1.602 \times 10^{-19})^4 (9.11 \times 10^{-31})}{2(4\pi(8.854 \times 10^{-12})^2(1.055 \times 10^{-34})^2}\) \(E_{1} \approx -122.4\) eV So, the ground state energy of \({Li}^{2+}\) is approximately -122.4 eV. #c) Ground state energy of deuterium#
03

Calculating ground state energy of deuterium#

Deuterium is an isotope of hydrogen with a neutron and proton in the nucleus. However, Bohr's model only depends on the atomic number and not the mass. Since the atomic number of hydrogen is \(Z = 1\), Bohr's formula can be applied without any changes. So, we set \(n = 1\): \(E_{1} = -\dfrac{(1)^2 e^4 m_e}{2(4\pi\epsilon_o)^2\hbar^2 (1)^2}\) Now, calculate the ground state energy: \(E_{1} = -\dfrac{(1)^2 (1.602 \times 10^{-19})^4 (9.11 \times 10^{-31})}{2(4\pi(8.854 \times 10^{-12})^2(1.055 \times 10^{-34})^2}\) \(E_{1} \approx -13.6\) eV So, the ground state energy of deuterium is approximately -13.6 eV.

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Most popular questions from this chapter

What is the minimum potential difference applied to an x-ray tube if \(x\) -rays of wavelength \(0.250 \mathrm{nm}\) are produced?
Suppose that you have a glass tube filled with atomic hydrogen gas (H, not \(\mathrm{H}_{2}\) ). Assume that the atoms start out in their ground states. You illuminate the gas with monochromatic light of various wavelengths, ranging through the entire IR, visible, and UV parts of the spectrum. At some wavelengths, visible light is emitted from the \(\mathrm{H}\) atoms. (a) If there are two and only two visible wavelengths in the emitted light, what is the wavelength of the incident radiation? (b) What is the largest wavelength of incident radiation that causes the \(\mathrm{H}\) atoms to emit visible light? What wavelength(s) is/are emitted for incident radiation at that wavelength? (c) For what wavelengths of incident light are hydrogen ions \(\left(\mathrm{H}^{+}\right)\) formed?
Calculate, according to the Bohr model, the speed of the electron in the ground state of the hydrogen atom.
A thin aluminum target is illuminated with photons of wavelength \(\lambda\). A detector is placed at \(90.0^{\circ}\) to the direction of the incident photons. The scattered photons detected are found to have half the energy of the incident photons. (a) Find \(\lambda\). (b) What is the wavelength of backscattered photons (detector at \(\left.180^{\circ}\right) ?\) (c) What (if anything) would change if a copper target were used instead of an aluminum one?
Find the energy in electron-volts required to remove the remaining electron from a doubly ionized lithium (Li \(^{2+}\) ) atom.
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