The ionization energy of hydrogen atom in the ground state is 1312 kJ "mol"^(-1) . Calculate the wavelength of radiation emitted when the electron in hydrogen atom makes a transition from n = 2 state to n = 1 state (Planck’s constant, h = 6.626 xx 10^(-34) Js, velocity of light, c = 3 xx 10^8 m s^(-1), Avogadro’s constant, N_A = 6.0237 xx 10^23 "mol"^(-1) ).

The ionization energy of hydrogen atom in the ground state is 1312 kJ

1.	The ionization energy of hydrogen atom in the ground state is 1312 kJ "mol"^(-1) . Calculate the wavelength of radiation emitted when the electron in hydrogen atom makes a transition from n = 2 state to n = 1 state (Planck’s constant, h = 6.626 xx 10^(-34) Js, velocity of light, c = 3 xx 10^8 m s^(-1), Avogadro’s constant, N_A = 6.0237 xx 10^23 "mol"^(-1) ).
Answer» Solution :I.E. of hydrogen atom in the ground state `= 1312 kJ "mol"^(-1)` Energy of hydrogen atom in the first orbit `(E_1) = -I.E = -1312 kJ "mol"^(-1)` Energy of hydrogen atom in the nth orbit `(E_n) = (-1312 )/(n^2) kJ "mol"^(-1)` Energy of hydrogen atom in the SECOND orbit `(E_2) = - (1312)/(2^2) = -328 kJ "mol"^(-1)` `Delta E = E_2 - E_1 = [-328 - (-1312)] kJ = 984 kJ "mol"^(-1)` Energy released per atom` = (Delta E)/(N) = (984 XX 10^3 J//"atom")/(6.023 xx 10^23) ` ` (Delta E)/(N) = hv = H c/lamda ,therefore lamda = (Nh_1 c)/(Delta E)` ` therefore lamda = (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1) xx 6.0237 xx 10^23)/(984 xx 10^3 J) = 1.2 xx 10^(-7) m`