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stepan [7]
3 years ago
5

g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t

he probability density for stimulated emission (or absorption). If N 2 such atoms are excited to energy level 2, determine the time constant for the decay of N 2 due to stimulated and spontaneous emission. How many photons (rather than 1000) should be present so that the decay rate due to stimulated emission equals that due to spontaneous emission?
Physics
1 answer:
Schach [20]3 years ago
7 0

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

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gladu [14]

Answer:

a)    A = 0.603 m , b) a = 165.8 m / s² , c)  F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

    Em₀ = K = ½ m v²

End Point with fully compressed spring

    Em_{f} = K_{e} = ½ k x²

    Emo = Em_{f}

    ½ m v² = ½ k x²

    x = √(m / k)    v

    x = √ (2.00 / 550)   10.0

    x = 0.603 m

This is the maximum compression corresponding to the range of motion

     A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

    F = m a

    k x = ma

    a = k / m x

    a = 550 / 2.00 0.603

    a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

    F = k x

    F = 550 0.603

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5 0
3 years ago
A car is traveling in a circular curve on a horizontal road at constant speed. Name the force that is responsible for causing it
boyakko [2]
The correct answer is
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Answer: A

Explanation:

3 0
2 years ago
A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are pl
Sonbull [250]

Answer:

A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

Explanation:

Parameters given:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between both rods = 4.5cm = 0.045

A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

Where

E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

E₁ = kQ/0.01²

E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²

E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]

E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]

E = 9 * 10⁹ * 15.143* 10⁻⁵

E = 1.36 * 10⁶N/C

B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.02²

E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²

E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]

E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]

E = 9 * 10⁹ * 5.74 * 10⁻⁵

E = 5.17 * 10⁵N/C

C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.03²

E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²

E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]

E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]

E = 9 * 10⁹ * 7.77 * 10⁻⁵

E = 6.993 * 10⁵N/C

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4 years ago
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Explanation:

7 0
3 years ago
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