shortest wavelengths in balmer series of Li2+
Answers
Answered by
1
Answer:
The shortest wavelength of the light emitted is when the electron jumps from Infinity to 3. For Li²⁺, λ = 9/R × 3² m. = 81/R m. This is the Required Answer.
Explanation:
HOPE that helps you
Answered by
0
Answer:
no
Explanation:
yes I think like that so answer is no
Similar questions