Solve this:


2. In an isolated Li2+. electron jumps from n = 2 to n = 1 . What is the recoil momentum of the ion (in kg- m/s)
 
(1) 5.44 x 10-27            (2) 48.96 x 10-27 
(3) 16.32 x 10-27           (4) 8.16 x 10-27

Dear Student ,

Energy of the photon emitted = E= E1-E2E = 13.611-14 eVE= 13.6×0.75 = 10.2 eVMomentum of the emitted photon = P = EC = 10.2×1.6×10-193×108 =5.44×10-27 NsThis momentum of the photon will be the recoiled momentum of the Li2+ atom.

Regards

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