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1 year ago

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]1 year 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

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A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

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3 years ago

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Inga [223]

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