The radius of the first orbit of hydrogen is r _ H , and the energy in the ground state is - 13.6 eV . Considering a ⁻ -particle with a mass 207 m_e revolving round a proton as in hydrogen atom, the energy and radius of proton and ⁻ - 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=- m e^4 8 ₀^2 h^2 1 n^2 Obviously E_n m array llll & & E_ E_e & = m_ m_e & E_ & = m_ m_e E_e array Ground state energy of a proton in hydrogen atom, aligned E_ & =-13.6 207 m_e m_e eV & =-13.6 207 eV aligned ( m_ =207 m_e . , where m_e is the mass of electron ) We know that r= ₀ h^2 n^2 207 m_ e e ^2 For ground state (n-1) for proton, we have r_ = ₀ h^2 207 m_e e^2 But ₀ h^2 m_ e e^2 = ground state radius of hydrogen atom aligned ₀ h^2 m_e e^2 & =r_ H r_ & = r_ H 207 aligned