Calculate the wavelength and energy of radiation for the elctronic transition form infinity to ground state for one H-atom . Given ` e_1 =- 13. 6 eV ( 1 1 eV = 1,. 6 xx 10^(-19) J)`.

Explanation

Given : `n_1 = 1, n_2 = infty`
` therefore Delta E = E_(infty) - E_1 =0- ( - 13.6 ) = 13.6eV`
` = 13.6 xx 1.602 xx 10^(-19) J = 217.9 xx 10^(-20)J`
Also `Delta E = (hc)/( lambda)`
` lambda= ( 6.626 xx 10^(-34 )xx 3.0 xx 10^8)/( 217 .9xx 10^(-20)) = 9. 12 xx 10^(-8) m`
`= 912xx10^(-10) m= 912Å`.