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3 years ago

What is spontaneous emission?

Physics

2 answers:

Oxana [17]3 years ago

5 0

Answer:

Spontaneous emission is a quantum effect, which in a semiclassical picture can be described as an emission which is stimulated by vacuum noise.

Irina-Kira [14]3 years ago

4 0

Spontaneous emission is the process in which a quantum mechanical system transits from an excited energy state to a lower energy state and emits a quantized amount of energy in the form of a photon. definition from wikipedia

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A device known as an optical resonator is used in lasers. An optical resonator consists of an arrangement of mirrors that reflec

polet [3.4K]

Answer:

A. The resonator behaves as a wave guide (a hollow pipe used as a transmission line). The characteristics of the pipe depend on the type of the wave to be transmitted.

4 0

3 years ago

A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre

Paraphin [41]

Answer: 80J

Explanation:

According to the first principle of thermodynamics:

"Energy is not created, nor destroyed, but it is conserved."

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy $U$ , which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

$\Delta U=Q+W$ (1)

Where:

$\Delta U$ is the variation in the internal (thermal) energy of the system (the value we want to find)

$Q=-100J$ is the heat transferred out of the gas (that is why it is negative)

$W$ is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

$W=-P \Delta V$ (2)

Where:

$P=450kPa=450(10)^{3}Pa$ is the constant pressure of the gas

$\Delta V=V_{f}-V_{i}$ is the variation in volume of the gas

In this case the initial volume is $V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}$ and the final volume is $V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}$ .

This means:

$\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}$ (3)

Substituting (3) in (2):

$W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})$ (4)

$W=180J$ (5)

Substituting (5) in (1):

$\Delta U=-100J+180J$ (6)

Finally:

$\Delta U=80J$ This is the change in thermal energy in the compression process.

8 0

3 years ago

Are we the same age as the universe because matter cannot be created nor destroyed

wlad13 [49]

No. We aren't the same age as the universe.

4 0

4 years ago

Read 2 more answers

Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t

faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;