Question

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

Answer 1

The recoil momentum of the ion is option (1): 5.44 x 10-27 Ns

Explanation:

• As we know that the energy formula is;

Energy of the photon emitted = E = E1 − E2

• Now put the values

E = 13.6 ( 1/1 - 1/4 )eV

E  = 13.6×0.75

   = 10.2 eV

• Now find the momentum:

Momentum of the emitted photon = P = EC

                                                         = 10.2×1.6×10^−19 / 3×108

                                                         = 5.44×10^−27 Ns

Hence this momentum of photon will be the recoiled momentum of the Li2+ ion.