Why is it difficult to accurately represent ammonia and methane on paper?
1 answer:
You might be interested in
The order of frequency is d < a < c < b
eV
where z = atomic mass number
n = energy level
For hydrogen z = 1
Therefore, Energy for n = 1
eV
= -13.6 eV
for n = 2
eV
= -3.40 eV
for n = 3
eV
= -1.51 eV
for n = 4
eV
= -0.85 eV
for n = 5
eV
= -0.544 eV
n = 2 to n = 4 (absorption)
ΔE = E4 - E2 = -0.85 - (-3.40) = 2.55 eV
n = 2 to n = 1 (emission)
ΔE = E1 - E2 = -13.6 - (-3.40) = -10.2eV
The negative sign indicates that emission will take place.
n = 2 to n = 5 (absorption)
ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV
n = 4 to n = 3 (emission)
ΔE = E3 - E4 = -1.51 - (-0.85) = -0.66 eV
We know that
E = h * υ
Therefore, Energy is proportional to frequency.
So increasing the order of energy is
E4 < E1 < E3 < E2
order of frequency is
d < a < c < b
For more information click on the link below:
# SPJ4