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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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A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th
zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

        v = ?

        u = 0 m/s

        a = 4.0 m/s²

         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

        a = -2.0 m/s²

         t = 4 s

         v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0
2 years ago
Rachel is helping her younger brother replace a broken part in his toy ambulance. This part is responsible for converting electr
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2 years ago
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On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the
Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

Learn more about distance and speed:

brainly.com/question/8893949

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4 0
2 years ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

5 0
2 years ago
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