JEE Advanced Model Paper 12 with solutions for online practice

In ground state , U1 = - 27.2 eV , K1 = 13.6 eV and i 1st excited state
U2 = - 6.8 eV and K2 = 3.4 eV
∴ kinetic energy decreases by 10.2eV
Potential energy increases by 20.4eV and increase of angular momentum =

h

2π

=

6.62

6.28

× 10−34 = 1.05× 10−34 J - S