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Ann [662]
3 years ago
11

Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of t

hese photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?
Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

4.5 x 10¹⁴ Hz

666.7 nm

1.8 x 10⁵ J

The color of the emitted light is red

Explanation:

E = energy of photons of light = 2.961 x 10⁻¹⁹ J

f = frequency of the photon

Energy of photons is given as

E = h f

2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f

f = 4.5 x 10¹⁴ Hz

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of photon

Using the equation

c = f λ

3 x 10⁸ = (4.5 x 10¹⁴) λ

λ = 0.6667 x 10⁻⁶ m

λ = 666.7 x 10⁻⁹ m

λ = 666.7 nm

n = number of photons in 1 mole = 6.023 x 10²³

U = energy of 1 mole of photons

Energy of 1 mole of photons is given as

U = n E

U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)

U = 1.8 x 10⁵ J

The color of the emitted light is red

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
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Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

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             v_{cm } = √(4/3 gh)

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             v_{cm }^ = 6.43 m / s

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