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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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