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The ionization energy of hydrogen atom i...

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)` ).

Text Solution

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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`
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