| A solution containing 4.48 ppm KMnO4 exhibits
1 answer:
Answer:
Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹
Explanation:
First convert Concentration from ppm inM or mol/l
⇒ Molar mass of KMnO₄ = 158.03 g
⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l
⇒ Molarity = = 2.83 x 10⁻⁵ molar
Absorbance (A) = - log(T) ( T = % transmittance)
= - log(0.859)
= 0.06
According to Lambert Beer's law
ε =
or, ε =
or, ε = 2120.14 cm⁻¹M⁻¹
Where
ε = Molar absorptivity
A = absorbance
C = Molar concentration of KMnO₄ solution
l = length
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The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
Answer:
Hence the order of the transition will be : d < a < c < b
Explanation:
where,
= energy of
orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy of n = 3 in an hydrogen atom:
Energy of n = 4 in an hydrogen atom:
Energy of n = 5 in an hydrogen atom:
a) n = 2 to n = 4 (absorption)
b) n = 2 to n = 1 (emission)
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
d) n = 4 to n = 3 (emission)
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
h = Planck's constant
frequency of the wave
So, the increasing order of magnitude of the energy difference :
And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :
: d < a < c < b