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solong [7]
4 years ago
9

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

ed by an electron beam so that there are some objects in each excited state, what are the energies of the photons that will be emitted?
Physics
1 answer:
gogolik [260]4 years ago
8 0

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

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enyata [817]

corrected question:The heavyweight boxing champion of the world punches a sheet of paper in midair, bringing it from rest up to a speed of 26.5 m/s in 0.044 s . The mass of the paper is 0.003 kg. Part A Find the force of the punch on the paper

Answer:

Force=1.8N

Explanation:

Newtons third law states that in every action there is equal and opposite reaction.

The force of the punch will be the force that moves the paper by a speed of 26.5m/s.

F=ma

F=m\frac{v-u}{t}

m=0.003kg , v=26.5m/s  u=0(the paper is punched from rest) t=0.044s

F=0.003*\frac{26.5-0.044}{0.044}

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4 0
3 years ago
A solid cylinder (1 =1/2mr2 ) with a mass of 4.83 kg and a radius of 0.057 m starts
netineya [11]

Answer:

v = 7.32 m/s

Explanation:

The potential energy will convert to kinetic energy

        ½Iω² + ½mv² = mgh

             Iω² +  mv² = 2mgh

(½mR²)(v/R)² + mv² = 2mgh

           ½mv² + mv² = 2mgh

                 ½v² + v² = 2gh

                      3v²/2 = 2gh

                           v² = 4gh/3

                           v² = 4(9.81)(4.10)/3

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3 years ago
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Ostrovityanka [42]

Answer:

Walking burns up more energy,1740000J

Explanation:

Given that the displacement is 5.0km, and running at 10km/h and uses, walking at 3km/hr and uses 290watts:

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700watts=700j/s,d=5000m,v=\frac{25}{9}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{25}{9}}\times 700j/s\\\\=1260000J

Energy consumption for walking is calculated as:

290watts=290j/s,d=5000m,v=\frac{5}{6}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{5}{6}}\times 290j/s\\\\=1740000J

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5 0
3 years ago
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IrinaVladis [17]
This is the same question that I just answered.

Have present the definition of acceleration:

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a and v are in bold to mean that they are vectors.

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A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

That body is not changing either direction or speed so its motion is not accelerated but uniform.

4) a body standing still : FALSE.

That body is not changind either direction or speed.

5) a body traveling at a constant speed and changing direction: CORRECT.

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