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Varvara68
3 months 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:
Ostrovityanka [3.2K]3 months ago
4 0

Answer:

(1) En to n-1 = 0.55 eV

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

(3) ninit =4

(4) L =483.676 ×10^-11 nm

(5) λlongest= 1773.33 nm

Explanation:

The comprehensive details regarding the answer are provided in the attached files.

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The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a
Sav [3153]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0
3 months ago
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
serg [3582]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: The speed of a wave on a string under tension can be determined using the following:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

4 0
3 months ago
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