Question

Calculate the ratio of the kinetic energy for the n = 2 electron for the Li atom to that of Be^+ ion?

(a) \(\frac {9}{16}\)

(b) \(\frac {3}{4}\)

(c) 1

(d) \(\frac {1}{2}\)

I got this question in exam.

The doubt is from Atomic Spectra in section Atoms of Physics – Class 12

Answer 1

The correct choice is (a) \(\frac {9}{16}\)

The explanation: \(\frac {KE_{Li}}{KE_{Be}} = \big [ \frac {(\frac {Z_{Li}}{2} )}{(\frac {Z_{Be}}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \big [ \frac {(\frac {3}{2} )}{(\frac {4}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \frac {9}{16}\).