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IrinaK [193]
3 years ago
7

An excited atom decays to its ground state and emits a photon of green light. If instead the atom decays to an intermediate stat

e, then the light emitted could be
Physics
1 answer:
scoundrel [369]3 years ago
8 0

Answer:

the light emitting must be of greater wavelength

Explanation:

For this exercise we must use the Planck equation

             E = h f

And the speed of light

            c = λ f

            f = c / λ

We replace

            E = h c / λ

The wavelength of the green light is of the order of 500 nm, let's calculate the energy

          E = 6.63 10⁻³⁴  3 10⁸ /λ

          E = 1,989 10⁻²⁵ /λ

          λ = 500 nm = 500 10⁻⁹ m

          E = 1,989 10⁻²⁵ / 500 10⁻⁹

          E = 3,978 10⁻¹⁹ J

That is the energy of the transition for a transition is an intermediate state the energy must be less, this implies that the wavelength must increase. For the explicit case of a state with half of this energy

            E_{int} = E / 2

             E_{int} = 3,978 10⁻¹⁹ / 2 = 1,989 10⁻¹⁹

Let's clear and calculate

           λ = h c / E

           λ = 1,989 10⁻²⁵ / 1,989 10⁻¹⁹

           λ = 1 10⁻⁶ m

Let's reduce to nm

          λ = 1000 nm

This wavelength is in the infrared region

the light emitting must be of greater wavelength

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Answer:

J = 1800 kg-m/s

Explanation:

Given that,

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We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,

J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s

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4 0
3 years ago
Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se
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Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω

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Help me pleaseee, it’s due today: Two people push on a large gate as shown on the view from above in the diagram. If the moment
Sladkaya [172]

Answer:

the angular acceleration of the gate is approximately 1.61  \frac{rad}{s^2}

Explanation:

Recall the formula that connects the net torque with the moment of inertia of a rotating object about its axis of rotation, and the angular acceleration (similar to Newton's second law with net force, mass, and linear acceleration):

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\sum \tau_1=I\,\alpha\\145\,\,N\.m=(90\,\,kg\,m^2)\,\alpha\\\alpha= \frac{145}{90} \frac{rad}{s^2} = 1.61\, \frac{rad}{s^2}

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