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vlada-n [284]
1 year ago
6

Learning Goal: To practice Problem-Solving Strategy 40.1 for quantum mechanics problems. Suppose a particle of mass m is confine

d to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box?

Physics
1 answer:
satela [25.4K]1 year ago
7 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.
OverLord2011 [107]

Answer is given below

Explanation:

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2 years ago
he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold
bekas [8.4K]

Answer:

Relative population is  2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

\frac{N_{1} }{N_{2} } =  \frac{g_{1}e^{\frac{-E_{1} }{KT} }  }{g_{2}e^{\frac{-E_{2} }{KT} }}

\frac{N_{1} }{N_{2} } = \frac{g_{1}e^{\frac{-(E_{1}-E_{2})  }{KT} }  }{g_{2}}

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

\frac{N_{1} }{N_{2} } = \frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} }  }{3}

\frac{N_{1} }{N_{2} } = 3.4 x 10⁹

\frac{N_{2} }{N_{1} } =  2.94 x 10⁻¹⁰

5 0
2 years ago
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Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

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Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
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Answer:

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