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The electronin a givenBohr orbit E_(n) = - 1.54 eV. Calculate (i) its kinetic energy (ii) potential energy and (iii) wavelenghtof light emitted , whenthe electron makes a transition to thegroundstate . Groundstate energyis- 13.6e V |
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Answer» Solution :(i) As total energy in givenBohr orbit `E_(n) =- 1.5 EV` `THEREFORE ` Kineticenergyof eletron `K_(n) = - E_(n)= + 1.5 eV` (II) The potential energyof electron `U_(n) = 2 E_(n) = - 3.0 eV` (iii) If electron makes a transition from ` - 1.5 eV`state to groundstate having energy ` - 13.6 eV`, theenergy of radiated photon. `E = - 1.5 -(-13.6) eV = 12.1 EC = 12.1 xx 1.60 xx 10^(19)J` As `E =hv = (hv)/(lambda) `, hencewavelenght of lightemitted is equal to `lambda = (HC)/(E) = (6.63xx 10^(-34) xx 3 xx 10^(8))/(12.1 xx 1.60 xx 10^(-19)) = 1.027 xx 10^(-7) m = 102.7` nm. `lambda =(hc)/(E) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(12.1 xx 1.60xx 10^(-19)) = 1.027 xx 10^(-7) m = 102.7 nm` |
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