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mariarad [96]
2 years ago
5

what is the frequency of the photon emitted by a hydrogen atom when its electron drops from the n = 6 to n = 4 energy level?

Physics
1 answer:
Gemiola [76]2 years ago
4 0

The frequency of the photon emitted by a hydrogen atom is   26.2*10^-^4 hz

<h3>What are Photons?</h3>

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

In the case of light, the frequency, symbolized by the Greek letter nu (ν), of any wave equals the speed of light, c, divided by the wavelength λ:

v = c/λ

Since the wavelength λ is in the bottom of the fraction, the frequency is inversely proportional to the wavelength.

v = c/λ

v= \frac{3*10^8}{434*10^-^9} = 6.91*10^1^4 Hz

By using Rydberg's formula,

1/λ = Rz^2(\frac{1}{n_1^2}  - \frac{1}{n_2^2})

1/λ = 109677 *(\frac{1}{4^2} - \frac{1}{6^2}) = 3808.22 cm^-^1

λ  = 26.2*10^-^4 hz

26.2*10^-^4 hz is  frequency of the photon emitted by a hydrogen atom.

Learn more about Photons here:brainly.com/question/20912241

#SPJ1

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4 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

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3 years ago
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8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o
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Answer:

The length of line is 78 cm or 0.78 m.

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initial reading 2 mark

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Answer:

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