Answer
a) $r_1=2.57\times 10^{-3}m$
b) less than
c) $\lambda=3.17\times 10^{-9}m=3.17nm$
Work Step by Step
(a) We know that
$r_1=\frac{h^2}{4\pi^2mke^2}$
We plug in the known values to obtain:
$r_1=\frac{(6.63\times 10^{-34})^2}{(4)(3.14)^2(207)(9.11\times 10^{-31})(9\times 10^9)(1.6\times 10^{-19})^2}$
$\implies r_1=2.57\times 10^{-3}m$
(b) We know that
$\frac{1}{\lambda}=(\frac{2\pi^2 m_{\mu}k^2e^4}{h^3c})(\frac{1}{n_f}^2-\frac{1}{n_i^2})$
This equation shows that wavelength is inversely proportional to the mass of the particle. As the mass of the muonium is greater than the mass of the electron, therefore the wavelength is smaller.
(c) We know that
$\lambda=\frac{1}{(207R)(\frac{1}{n_f}^2-\frac{1}{n_i}^2)}$
We plug in the known values to obtain:
$\lambda=\frac{1}{(207)(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{3^2})}$
$\lambda=3.17\times 10^{-9}m=3.17nm$
