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mariarad [96]
1 year 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]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

#SPJ1

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A wrecking ball is suspended by two cables as shown below. If the tension in cable 2 is 12 000 N, what is the weight of the wrec
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5 0
3 years ago
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
Flauer [41]

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = \sqrt{\frac{k}{m} }

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =\frac{dx}{dt}

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = \frac{dv}{dt}

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

5 0
2 years ago
At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev
olganol [36]

Answer:

the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

<em>v₂ = 73.4 m/s </em>

<em></em>

Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

5 0
3 years ago
Name nine elements that have been around so long that we don't even know when they
lora16 [44]

Answer:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

Explanation:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

-  (These are the most longest elements that have been around very long )

7 0
3 years ago
Read 2 more answers
Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
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