The Bohr model currently predicts the main energy levels not only for atomic hydrogen but also for other “one-electron” atoms where all but one of the atomic electrons has been removed, such as in ${\mathbf{He}}^{\mathbf{+}}$ (one electron removed) or (two electrons removed) ${\mathbf{Li}}^{\mathbf{+}\mathbf{+}}$. (a) Predict the energy levels in for a system consisting of a nucleus containing protons and just one electron. You need no recapitulate the entire derivation for the Bohr model, but do explain the changes you have to make to take into account the factor . (b) The negative muon$\left({\mu }^{-}\right)$ behaves like a heavy electron, with the same charge as the electron but with a mass 207 times as large as the electron mass. As a moving ${\mathbf{\mu }}^{\mathbf{-}}$ comes to rest in matter, it tends to knock electrons out of atoms and settle down onto a nucleus to form a “one-muon” atom. For a system consisting of a lead nucleus ( ${\mathbf{\text{Pb}}}^{\mathbf{\text{208}}}$has 82 protons and 126 neutrons) and just one negative muon, predict the energy in of a photon emitted in a transition from the first excited state to the ground state. The high-energy photons emitted by transitions between energy levels in such “muonic atoms” are easily observed in experiments with muons. (c) Calculate the radius of the smallest Bohr orbit for a ${\mathbf{\mu }}^{-}$ bound to a lead nucleus ( ${\mathbf{\text{Pb}}}^{\mathbf{\text{208}}}$has 82 protons and 126 neutrons). Compare with the approximate radius of the lead nucleus (remember that the radius of a proton or neutron is about $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathbf{\text{m}}$, and the nucleons are packed closely together in the nucleus).

Comments: This analysis in terms of the simple Bohr model hints at the results of a full quantum-mechanical analysis, which shows that in the ground state of the lead-muon system there is a rather high probability for finding the muon inside the lead nucleus. Nothing in quantum mechanics forbids this penetration, especially since the muon does not participate in the strong intersection. Electrons in an atom can also be found inside the nucleus, but the probability is very low, because on average the electrons are very far from the nucleus, unlike the muon.

The eventual fate of the ${\mathbf{\mu }}^{\mathbf{-}}$ in a muonic atom is that it either decays into an electron, neutrino, and antineutrino, or it reacts through the weak interaction with a proton in the nucleus to produce a neutron and a neutrino. This “muon capture” reaction is more likely if the probability is high for the muon to be found inside the nucleus, as is the case with heavy nuclei such as lead.

Step by step solution

01

 Definition of angular Acceleration

The rate of change of angular velocity with respect to time is known as the Angular acceleration.

02

Derive the expression of energy level

In hydrogen atom there is coulomb electrostatic attractive force between one proton and one electron. According to Bohr model, the energy levels for the orbital states of a hydrogen atom is given as

${\mathrm{E}}_{\mathrm{N}}=\frac{{\mathrm{ke}}^2}{2{\mathrm{r}}_{\mathrm{N}}}\text{eV}······\left(1\right)$

Radius of ${\mathrm{N}}^{\mathrm{th}}$orbit

Charge of the electron$\left(1.6\times {10}^{-19}\text{C}\right)$

In hydrogen like atoms here is coulomb electrostatic attractive force betweenprotons and one electron provide the centripetal force. Hence, we need to multiply equation (1) with.

The energy levels can be expressed as

${\mathrm{E}}_{\mathrm{N}}=\frac{{\mathrm{kZe}}^2}{2{\mathrm{r}}_{\mathrm{N}}}\text{eV}······\left(2\right)$

Hence,

03

Step 3(a):  Predict the energy levels

Mass of muon,${\mathrm{m}}_{\mathrm{\mu }}=207{\mathrm{m}}_{\mathrm{e}}$

Mass of the electron,${\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31}\text{kg}$

For lead atomic number, $\mathrm{Z}=82$

For ground state,$\mathrm{Z}=82$

The radius of the hydrogen like atom is give as

$\begin{array}{rcl}{r}_{\mathrm{n}}& =& \frac{{\mathrm{kZe}}^2}{2\left(\frac{{\mathrm{N}}^2{\mathrm{h}}^2}{{\mathrm{mkZe}}^2}\right)}······\left(4\right)\\ & =& \frac{{\mathrm{Z}}^2{\mathrm{k}}^2\mathrm{m}{}_{\mathrm{e}}{\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}\\ & & \end{array}$

For muonic atom, the mass of electron in equation (4) is replaced by mass of muon. And it can be expressed as

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{N}}& =& -\frac{{\mathrm{Z}}^2{\mathrm{k}}^2\mathrm{m}{}_{\mathrm{\mu }}{\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}\\ & =& -\frac{{\mathrm{Z}}^2{\mathrm{k}}^2\left(207{\mathrm{m}}_{\mathrm{e}}\right){\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}······\left(5\right)\\ & =& -207\frac{{\mathrm{Z}}^2{\mathrm{k}}^2{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}\\ & & \end{array}$

On substituting all values in equation (5), the ground state energy of the muonic lead atom can be calculated as

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{N}-1}& =& -207\frac{{\mathrm{Z}}^2{\mathrm{k}}^2\mathrm{m}{}_{\mathrm{e}}{\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}\\ & =& -\left(207\right)\frac{{\left(82\right)}^2\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(9.1\times {10}^{-31}\text{kg}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^4}{2{\left(1\right)}^2{\left(1.055\times {10}^{-34}\text{J}·\text{s}\right)}^2}\\ & =& 1.9\times {10}^7\text{eV}\\ & & \end{array}$

The first excited energy state $\left(\mathrm{N}=2\right)$ of muonic lead atom can be calculated as

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{N}=1}& =& -207\frac{{\mathrm{Z}}^2{\mathrm{k}}^2{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^4}{2{\mathrm{N}}^2{\mathrm{h}}^2}\\ & =& -\left(207\right)\frac{{\left(82\right)}^2\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^2\right)\left(9.1\times {10}^{-31}\text{kg}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^4}{2{\left(2\right)}^2{\left(1.055\times {10}^{-34}\text{J}·\text{s}\right)}^2}\\ & =& -4.719\times {10}^6\text{eV}\\ & & \end{array}$

Hence, the first excited energy state $\left(\mathrm{N}=2\right)$ of muonic lead atom is $4.719\times {10}^6\text{eV}$.

04

Step 4(b): Find the energy of the photon emitted between the first excited state and ground state energy state

The energy of the photon emitted between the first excited state and ground state energy state is can be calculated as
$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{photon}}& =& {\mathrm{E}}_{\mathrm{N}=2}-{\mathrm{E}}_{\mathrm{N}=1}\\ & =& -\left(4.719\times {10}^6\text{eV}\right)-\left(-1.9\times {10}^7\text{eV}\right)\\ & =& 1.43\times {10}^7\text{eV}\\ & & \end{array}$

Hence, the energy of the photon emitted between the first excited state and ground state energy state is $1.43\times {10}^7\text{eV}$.

05

Step 5(c): Find the radius of the muonic lead atom

The radius of the muonic lead atom can be calculated as

$\begin{array}{rcl}{\mathrm{r}}_{\mathrm{N}}& =& \frac{{\mathrm{N}}^2{\mathrm{h}}^2}{{\mathrm{m}}_{\mathrm{\mu }}{\mathrm{kZe}}^2}\\ & =& \frac{{\mathrm{N}}^2{\mathrm{h}}^2}{207{\mathrm{m}}_{\mathrm{e}}{\mathrm{kZe}}^2}\\ & =& \frac{{\left(1\right)}^2{\left(1.055\times {10}^{-34}\text{J}·\text{s}\right)}^2}{\left(207\right)\left(9.1\times {10}^{-31}\text{kg}\right)\left(9\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^2\right)\left(82\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}\\ & =& 3.13\times {10}^{-15}\text{m}\\ & & \end{array}$

Hence, The radius of the muonic lead atom is $3.13\times {10}^{-15}\text{m}$ .

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