The energy of an electron in an orbit of hydrogen like ion with an orbit radius of (52.9 pm ) in ( J ) is (ground state energy of electron in hydrogen atom is (=-2.18 10⁻¹⁸ ~J ) )

Step-by-step solution

Given, Orbit radius (=52.9 pm ) Ground state energy of electron in hydrogen atom ( aligned & =-2.18 10⁻¹⁸ ~J r_n & =r₀ n^2 Z 52.9 pm & =52.9 pm n^2 Z n^2 Z =1 or n^2=Z aligned ) From energy, ( aligned & E_n=E₀ Z^2 n^2 & E_n=-2.18 10⁻¹⁸ n^4 n^2 & E_n=-2.18 10⁻¹⁸ n^2 & ( n^2=Z ) & E_n=4 -2.18 10⁻¹⁸=-8.72 10⁻¹⁸ ~J aligned ) If (n=2 ), from given option, then (E_n=4 -2.18 10⁻¹⁸=-8.72 10⁻¹⁸ ~J ) Thus, option (3) is correct.