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vivado [14]
3 years ago
7

The Balmer Series of lines in the hydrogen atom was very important in the history of physics and astronomy. What made it so impo

rtant?
A) It helped Rutherford explain the workings of the atomic nucleus.
B) Balmer's experiments with these lines helped prove Wien's Law.
C) When electrons change levels to produce the Balmer series, the emitted light is in the visible part of the spectrum.
D) The Balmer series of lines can only be observed from outer space; so the competition to see these lines led to the space race.
E) Mr. Balmer never got the Nobel prize, which made everyone in the field of spectroscopy so angry, they worked extra hard.
Chemistry
1 answer:
True [87]3 years ago
3 0

Answer:

C) When electrons change levels to produce the Balmer series, the emitted light is in the visible part of the spectrum

Explanation:

The Balmer Series of lines in the hydrogen atom was very important in the history of physics and astronomy. What made it so important?

A) It helped Rutherford explain the workings of the atomic nucleus.

B) Balmer's experiments with these lines helped prove Wien's Law.

C) When electrons change levels to produce the Balmer series, the emitted light is in the visible part of the spectrum.

D) The Balmer series of lines can only be observed from outer space; so the competition to see these lines led to the space race.

E) Mr. Balmer never got the Nobel prize, which made everyone in the field of spectroscopy so angry, they worked extra hard.

the most appropriate will be

C) When electrons change levels to produce the Balmer series, the emitted light is in the visible part of the spectrum

Balmer series of hydrogen lines. These lines are emitted when the electron in the hydrogen atom moves from the orbital n = 3 or greater shell down to the n = 2 shell/orbit. ... Energy is emitted from the atom when the electron leaps from one orbit to another closer to the nucleus.

Wiens law has to do with the radiation  of a black body.

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| A solution containing 4.48 ppm KMnO4 exhibits
Artist 52 [7]

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 = \frac{4.48 X10^{-3} }{158.03X 1(lit)} = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

                          = - log(0.859)

                          = 0.06

According to Lambert Beer's law

     

                 ε = \frac{A}{C X l}

      or,      ε = \frac{0.06}{2.83 X 10^{-5}X1 cm }

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

    ε = Molar absorptivity

    A = absorbance

    C = Molar concentration of KMnO₄ solution

     l = length  

6 0
3 years ago
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