Question

Atoms
11. The wavenumber of the energy emitted when electron jumps from fourth orbit to second
orbit in hydrogen atom is 20,397 cm-'. The wavenumber of energy for the same transition
in Het is
1) 81,988 cm
2) 40,944 cm-
3)20497 cm
4) 5.099cm-1​

Answer 1

Answer:

answer is 4)5.099cm-1

it is the answer of your question check it

Answer 2

Explanation:

z=atomic number.

for H z=1 and for He+ z=2 so just plug in the values in wavenumber formula. for solution check the photo. 1).81988 is the right answer.