Question

given that in h atom the transition energy for n=1 to n=2 rydberg state is 10.2ev the energ for same transition on be+3 is

Answer 1

Given that in the H- atom, the transition energy for n = 1 to n = 2 Rydberg states is 10.2eV, The energy for the same transition in Be3+Be3+ is :

A

20.4 eV

B

163.2 eV

C

30.6 eV

D

40.8 eV

Solution

E2=−13.6×164;E1=−13.6×16=−217.6E2=−13.6×164;E1=−13.6×16=−217.6
=−13.6×4=−64.4=−13.6×4=−64.4
E2−E1=−64.4+217.6=163.2