JEE Advanced Model Paper 9 with solutions for online practice

photon energy liberated from $$He^{\+} =

hc

λ

= hcRH × Z2[

1

12

−

1

22

] (for I line of lyman series)
= 6.626 × 10−27 × 3 × 1010 × 109678 × 22 ×

3

4

= 6.54 × 10−11 erg
This energy isused in liberating electron from H-atom from ground state
= 6.54 × 10−11 - 13.6 × 1.602 × 10−12=4.362×10^{-11}$$ erg = 1/2 mu2
∴ u = 3.09 × 108 cm sec−1