Answer:
a. see attachment
b. 2535ev
c. 4.40nm
Explanation:
Please kindly check attachment for the detailed and step by step solution of the given problem.
Following are solution to the given equation:
The muon's reduced mass is determined as follows:
[tex]\to \mu =\frac{mM}{m+M}=\frac{(106 Me \frac{V}{c^2})(938 Me \frac{V}{c^2})}{(106 Me \frac{V}{c^2})+(938 Me \frac{V}{c^2})}\\\\ \to \mu= 95.2 Me \frac{V}{c^2} \\\\[/tex]
The electron's charge is [tex]e=1.6 \times 10^{-19}\ C[/tex] Where the free space's permittivity is [tex]\varepsilon_0 = 8.85 \times 10^{-12} \ \frac{C^2}{N.m^2}[/tex]. So,
[tex]\to \frac{e^2}{4 \pi \varepsilon_0 } =1.44 \times 10^{-9} eV.m \\\\[/tex]
Inside the ground state, the lowest radius is calculated as follows:
[tex]\to a_0=\frac{4 \pi \varepsilon_0 h^2 }{\mu e^2}=\frac{h^2}{(\frac{e^2}{4\pi \varepsilon_0 }) \mu}\\\\[/tex]
[tex]= \frac{(6.58\times 10^{-16} eV.s)^2}{(1.44 \times 10^{-9}\ eV.m)(95.2 \times 10^6 \frac{eV}{c^2})} \\\\=3.15 \times 10^{-30} \ c^2 (\frac{3.00 \times 10^8}{c})^2 \\\\= 2.84 \times 10^{-13} \ m\\\\[/tex]
For point b:
The muon's binding energy in the ground state is
[tex]\to E=\frac{e^2}{8 \pi \varepsilon_0 a_0 }[/tex]
[tex]=\frac{1}{2a_0}(\frac{e^2}{4 \pi \varepsilon_0 })\\\\=\frac{(1.44 \times 10^{-9}\ eV.m)}{2(2.84 \times 10^{-13}\ m)} \\\\=\frac{1.44 \times 10^{-9}\times 10^{13} \ eV}{5. 68 } \\\\ = 2535\ eV\\\\[/tex]
For point c:
The wavelength for the first series (n=1) is,
[tex]\to \lambda =n^2 (\frac{hc}{E})[/tex]
[tex]= \frac{ (1)^2 (1240 eV .nm)}{2535\ eV} \\\\= \frac{ 1 \times 1240 eV .nm}{2535\ eV} \\\\= \frac{ 1240\ nm}{2535} \approx 0.49\ nm\\\\[/tex]
The wavelength for the second series (n=2) is,
[tex]\to \lambda= \frac{(n)^2 hc}{E}[/tex]
[tex]= \frac{(2)^2 (1240 eV .nm)}{2535\ eV}\\\\ = \frac{ 4 \times 1240 eV .nm}{2535\ eV}\\\\ = \frac{ 4960 nm}{2535\ eV}\ \approx 1.96 \ nm \\\\[/tex]
The wavelength for the third series (n=3) is,
[tex]\to \lambda =\frac{(n)^2 hc}{E}[/tex]
[tex]=\frac{(3)^2 (1240 eV.nm)}{2535 \ eV} \\\\ =\frac{ 9 \times 1240 \ nm}{2535 } \\\\= \frac{ 11160 \ nm}{2535 }\\\\ = 4.40\ nm \\\\[/tex]
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