Tasks regarding lasers

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Tasks Regarding Lasers
Question 1
Spontaneous emission is the process by which a quantum mechanical system like an atom or subatomic particles, transit from an excited energy state to an area of low energy state with an external agent (Corder et al. 22). It is found in LEDs and fluorescent tubes. While stimulated emission is the process where photons are used to produce other photons that have equal phase and wavelength as the parent. This process is found in the laser beam.
Question 3
n5s= e-(3.4 *10-19J)/(1.38*10-23 J/K)(e-98.7=7.1*10-42 n3p)
n5s = e −(3.4⋅10−19 J)/(1.38⋅10−23J/K)(500K) = e−417.4 = 2.7⋅10−21 n3p
n5s = e −(3.4⋅10−19 J)/(1.38⋅10−23J/K)(1200K) = e−19.6 = 2.1⋅10−11 n3p
at all the temperature levels, all the atoms are in their energized states, therefore, the first state for their transition emits 632.8nm radiation which is almost negligible. We can conclude that the ratio is directly proportional to the temperature.

Question 4
Three level laser cannot produce a continuous beam because the energy level of E1 is less than E2 and E3. Additionally, E2 is greater than E1 but less than E3. Therefore, the energy level of E2 lies between E3 and E1.
Question 9
In a laser, the mirror is used to reflect photons, to stimulate atoms to emit more photons. Additionally, mirrors make up the optical cavity, responsible for accelerating photons across the active material. The process duplicates more photons resulting in exponential growth of electromagnetic energy.

Moreover, the mirror plays the role of maintaining, generating and amplifying the laser beam through forming a reflective resonator.
Question 12
g (E) dE equals to the number densities of electrons state in cm-3 in the range of (E, E+dE),
It follows that,
gc (E) dE = mn*√{2mn*(E œ Ec)} / (π2h3)
gv (E) dE = mp*√{2mp*(Ev œ E)} / (π2h3)
From the Fermi function we find that the electrons will distribute in the following manner.
f (E) = 1/ (1 + exp (E œ Ef)/kT).
The interval (E, E+dE) has (E) g (E) dE electrons
In the case of a doped semiconductor, the position of Ef in relation to the band gap limits the availability of electrons or holes.
Through integration;
n = in exp (Ef – Ei)/kT
p = ni exp (Ei – Ef)/kT Where
ni = Nc exp (Ei – Ec)/kT
Nc = 2(2πmn*kT/h2)3/2 =” effective density of conduction band states ‘
Ei = Fermi level in the intrinsic case.
In the same manner for N v., therefore,
Np = ni2 at equilibrium
ni2 = Nc Nv exp (Ev – Ec)/kT = Nc Nv exp (-Eg)/kT
It follows that;
Ef – Ei = kT ln (n/ ni) = – kT ln (p/ ni)
~ kT ln (ND / ni) or – kT ln (NA / ni)
Neutron-type Proton Type. Therefore, N1=N2.
Work Cited
Corder, Christopher, Brian Arnold, and Harold Metcalf. “Laser cooling without spontaneous emission.” Physical review letters114.4 (2015): 043002.