The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d
Rydberg's formula :
,
where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from 
to 
and 
= 109677 is the Rydberg Constant. Here and represents the transitions.
(a) =2 to = infinity
= 109677/4 [since 1/infinity = 0] Therefore, 
= 4 / 109677 = 0.00003647 m
(b) =4 to = 20
= 6580.62
Therefore, = 1 / 6580.62 = 0.000152 m
(c) =3 to = 10
= 11089.56
Therefore, = 1 / 11089.56 = 0.00009 m
(d) =2 to = 1
= - 82257.75
Therefore, = 1 /82257.75 = - 0.0000121 m
[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]
So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.
Learn more about the Rydberg's formula athttps://brainly.com/question/14649374
#SPJ4
Answer: D, splitting water into hydrogen and oxygen is a chemical change.
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
To calculate the hybridization of 
, we use the equation:
![\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom (S) = 6
N = number of monovalent atoms bonded to central atom = 0
C = charge of cation = 0
A = charge of anion = 0
Putting values in above equation, we get:
![\text{Number of electron pair}=\frac{1}{2}[6]=3](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5B6%5D%3D3)
The number of electron pair around the central metal atom are 3. This means that the hybridization will be 
and the electronic geometry of the molecule will be trigonal planar.
Hence, the correct answer is Option D.
We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
