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The value of ground state energy of hydrogen atom is - 13.6 eV. (a) Find the energy required to move an electron from the ground state to the first excited state of the atom. (b) Determine (i) the kinetic energy, and (ii) the orbital radius in the first excited state of the atom. (Giventhe valueof Bohr radius= 0.53Å) |
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Answer» Solution :(a) `because` Ground state energy of electron (N = 1) is E = - 13.6 eV `therefore ` Energy for firstexcited state `(n =2) E_(2) =(-13.6)/(n^(2)) = (-13.6)/((2)^(2)) = - 3.4 eV` `therefore ` Energyrequired to be given to electron `Delta E = E_(2) - E_(1) = - 3.4 - (-13.6) = + 10.2 eV` (b) (i) Kineticenergyof electron in 1st excited state `K_(2) = -E_(2) = + 3.4 eV` (ii) Orbital RADIUS in first excited state `r_(2) = n^(2) a_(0) = (2)^(@) xx 0.53 Å = 2.12 Å` |
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