Question

46. The ratio of the energy required to remove electron
from second orbit and third orbit of He+ ion is
(1) 2:3
(2) 1:1
(3) 9:4
(4) 4:1​

Answer 1

Answer:

3)9:4

Explanation:

Energy directly proportional to z square and inversely proportional to n square

Z same here so cancer on each

Depends on n value it comes 9:4