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Harrizon [31]
2 years ago
14

An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate

ly 19 eV. (a) What is the quantum number n for the energy state that the electron occupies
Physics
1 answer:
Svet_ta [14]2 years ago
5 0

The quantum number of the energy state occupied by the electron is 10.

The given parameters;

  • <em>distance between the two rigid walls, L = 1.4 nm</em>
  • <em>energy of the electron, E = 19 eV</em>

The quantum number of the electron is calculated as follows;

E_n = \frac{h^2n^2}{8mL^2} \\\\

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
  • <em>m is mass of electron = 9.11 x 10⁻³¹ kg</em>
  • <em>E is the energy of the electron = 19 x 1.602 x 10⁻¹⁹ J</em>
  • <em>n is the quantum number of the energy state occupied by the electron.</em>

n^2 = \frac{8mL^2E_n  }{h^2} \\\\n = \sqrt{\frac{8(9.11\times 10^{-31} )(1.4 \times 10^{-9})^2 (19\times 1.602 \times 10^{-19})  }{(6.626\times 10^{-34})2} } \\\\n \approx 10

Thus, the quantum number of the energy state occupied by the electron is 10.

Learn more here:brainly.com/question/19426524

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Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

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Where:

μ: is the coefficient of friction

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g: is the gravity

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The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

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