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maks197457 [2]
2 years ago
15

An electron is trapped in a square well of unknown width, L. It starts in unknown energy level, n. When it falls to level n-1 it

emits a photon of wavelength λphoton = 2280 nm. When it falls from n-1 to n-2, it emits a photon of wavelength λphoton = 3192 nm.
1) What is the energy of the n to n-1 photon in eV?

En to n-1 =

2) What is the energy of the n-1 to n-2 photon in eV?

En-1 to n-2 =

3) What is the initial value of n?

ninitial =

4) What is the width, L, of the well in nm?

L =

5) What is the longest wavelength of light, λlongest, the well can absorb in nm?

λlongest =

Physics
1 answer:
Lady bird [3.3K]2 years ago
4 0

Answer:

(1) En to n-1 = 0.55 ev

(2) En-1 to n-2 = 0.389 ev

(3) ninitial =4

(4) L =483.676 ×10^-11 nm

(5) λlongest= 1773.33 nm

Explanation:

Detailed explanation of answer is given in the attached files.

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For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
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The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

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<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
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<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
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<span>r = v × t </span>
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<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
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<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
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Marysya12 [62]
<h2>Answer:</h2>

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