1.

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»

The CORRECT CHOICE is (a) \(\frac {9}{16}\)

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

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

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



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