The radius of the first orbit of hydrogen is r _ H , and the energy in the ground state is - 13.6eV . Considering a μ ^ - - particle with a mass 207 m _ e revolving around a proton as in hydrogen atom, the energy and

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Correct answer: A. - 13.6 × 207eV, (rH)/(207)

The total energy of n^ th orbit, E _ n = - me4 8 ε _ 0 ^ 2 h^ 2 · (1)/(n2) Obviously E _ n ∝ m ∴(Eμ)/(Ee) = (mμ)/(me) ⇒ E _ μ = (mμ)/(me) × E _ e Ground state energy of a proton in a hydrogen atom, E _ μ = - 13.6 × 207 m _ e me eV = - 13.6 × 207eV ( ∵ 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 = (rH)/(207)

The radius of the first orbit of hydrogen is r _ H , and the energy...

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