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solong [7]
3 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]3 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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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved
mr_godi [17]

Answer:

A

Explanation:

6 0
2 years ago
When two objects with electrical charges interact, which affect the strength of that interaction?
noname [10]
The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span> 
6 0
3 years ago
a person does 100 joules of work in pulling back the string of a bow. what will be the initial speed of a 0.5 kg arrow when it i
eduard

If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.

                 Kinetic energy = (1/2) (mass) (speed)²

                         100 J        = (1/2) (0.5 kg) (speed²)

Divide each side by 0.25 kg:    100 J / 0.25 kg = speed²

 [ joule ] = [ newton-meter ] = kg-m²/sec²

                                                100 kg-m²/sec² / 0.25 kg = speed²

                                                 400 m²/sec² = speed²

Take the square root of each side:     speed = √400 m/s

                                                                                20 m/s

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3 0
3 years ago
A process with a negative change in enthalpy and a negative change in entropy will generally be:________
valentina_108 [34]

A process with a negative change in enthalpy and a negative change in entropy will generally be: <u>spontaneous</u>.

<h3>Gibbs free energy:</h3>

Since the Gibbs free energy is a parameter that tells us whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) in this situation, we can describe it mathematically as:

                                      ΔG = ΔH - TΔS

Therefore, any process with a negative change in enthalpy and a positive change in entropy will be spontaneous. If the enthalpy and the entropy are both negative, the subtraction becomes always negative, for which the Gibbs free energy is also negative.

One of the most crucial thermodynamic functions for the characterization of a system is the Gibbs free energy. It influences results like the voltage of an electrochemical cell and the equilibrium constant for a reversible reaction, among others.

Learn more about spontaneous here:

brainly.com/question/16975806

#SPJ4

3 0
1 year ago
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