The radius of the first orbit of hydrogen is r H , and the energy in the ground state is - 13.6 eV . Considering a μ - - particle with a mass 207 m e revolving around a proton as in hydrogen atom, the energy and radius of proton and μ - -combination respectively in the first orbit are (assume nucleus to be stationary)

Step-by-step solution

The total energy of n th orbit, E n = - me 4 8 ε 0 2 h 2 · 1 n 2 Obviously E n ∝ m ∴ E μ E e = m μ m e ⇒ E μ = m μ m e × E e Ground state energy of a proton in a hydrogen atom, E μ = - 13.6 × 207 m e m e eV = - 13.6 × 207 eV ( ∵ m μ = 207 m e , where m e is the mass of an electron) We know that, r = ε 0 h 2 n 2 207 πm e e 2 For ground state n= 1 for proton, we have r μ = ε 0 h 2 207 πm e · e 2 But ε 0 h 2 πm e · e 2 = ground state radius of a hydrogen atom ε 0 h 2 πm e · e 2 = r H ∴ r μ = r H 207