Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

If a radiation corresponds to second line of Balmer series of  Li+2  ion knocked out electron from first excited state of H – atom then K.E of ejected electron in ev is___ 

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

answer is 19.55.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

E=13.6×32[122142]=13.6×2716 
Energy required to remove e  from 2nd orbit of H is  =13.6×12(141α2)=13.64
K.Eof  e=13.6×271613.64=19.55  ev

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon