Lasers: Important Solved Problems

 

ANNA UNIVERSITY SOLVED PROBLEMS

 

1. A cadmium sulphide photo detector crystal is irradiated over a receiving area of 4×10^‒6 m² by light of wavelength 0.4×10^‒6 m and intensity 200 Watt/m². Assuming that each quantum generates an electron hole pair, calculate the number of pairs generated per second.

Solution

Intensity per unit area = Number of photons × Energy of a photon

Number of photons = (Intensity × area) / Energy of photon (hv)

= 200×4×10^‒6 / {6.625×10^‒34×[ 3×10⁸ / 0.4×10⁻⁶]}

= 8×10^‒4 / 4.96875×10^‒19

Number of photons = 1.6100 × 10¹⁵

Since each photon generates an electron‒hole pair, the number of photons is equal to the number of electron‒hole pairs.

Number of electron hole pairs = 1.6100×10¹⁵

 

2. Calculate the Wavelength of radiation emitted by an LED made up of a semiconducting material with band gap energy 2.8 eV.

Solution

Given

E_g = 2.8 eV

= 2.8 × 1.602 × 10^‒19 J

= 4.4856 × 10^‒19 J

We know E_g = hv Joules

E_g = hc / λ

∴  λ = hc / E_g

= 6.625×10^‒34×3×10⁸ / 4.4856×10^‒19

λ = 4430.8 Å (Blue colour)

 

3. For InP laser diode, the wavelength of light emission is 1.55 μm. What is its band gap in eV?

Solution

Given λ = 1.55×10^‒6 m,

E_g = ?

Energy band gap E_g = hv Joules

E_g = hc / λ

E_g = 6.625×10^‒34×3×10⁸ / 1.55×10^‒6

E_g = 1.28225×10^‒19 Joules

E_g = 1.28255×10⁻¹⁹ / 1.6×10^‒19 eV

= 0.8014 eV

Energy band gap E_g = 0.8014 eV

 

4. Calculate the number of photons from green light of mercury (λ = 4961Å) requires to do one joule of work.

Solution

Given λ=4961×10^‒10m

We know E = hv = hc / λ

 = 6.625×10^‒34×3×10⁸ / 4961×10^‒10  Joules

E = 4.006×10^‒19 Joules

Number of photons required to do one Joule of work (N) = 1 Joules / 4.006×10^‒19 Joules

N = 2.4961×10¹⁸/m³

 

5. Calculate the long wavelength limit of a extrinsic semiconductor if the ionisation energy is 0.02 eV.

Solution

The ionization energy = 0.02 eV

We know E = hv=  hc / λ

E = hc / λ

λ_long = hc / E

λ_long = 6.625×10^‒34×3×10⁸ / 0.02×1.6×10^‒19

λ_long = 6.210 × 10^‒5 m

 

6. Calculate the wavelength of emission from GaAs semiconductor laser whose band gap energy is 1.44 eV (Planck's constant is 6.625×10^‒34 Js and charge of an electron is 1.6×10^‒19 C.

Given data

Band gap energy E_g = 1.44 eV (or) 1.44 × 1.6 × 10^‒19 Joules

Solution

We know Band gap energy (E_g) = hv = hc / λ  

(or) we can write λ = hc / E_g

λ = 6.625×10^‒34×3×10⁸ / 1.44×1.6×10^‒19

 λ = 8.6263 × 10^‒7 m

∴ Wavelength of GaAs laser = 8626.3 Å

 

7. For a semiconductor laser, the band gap is 0.8 e V. What is the wavelength of light emitted from it.

Given Data:

 E_g = 0.8 eV = 0.8 × 1.6×10^‒19 Joules

Formula:

Wavelength λ = hc / E_g

λ = 6.625×10^‒34×3×10⁸ / 0.8×1.6×10^‒19

∴ λ=1.5527×10^‒6m

The wavelength of light emitted = 1.5527 μm

 

8. Calculate how many photons are emitted in each minute in a He‒Ne laser source, which emits light at a wavelength of 6328Å. The output power of this source is 3mW.

Solution

The frequency of the photon emitted by the laser beam = c/λ

 = 3×10⁸ / 6328×10^‒10

v = 4.74 × 10¹⁴ Hz

We know that the energy of a photon E = hv

 (or) E = 6.625 × 10^‒34 × 4.74 × 10¹⁴

(or) E = 3.14×10^‒19 J

Energy emitted by the laser = 3mW = 3×10^‒3 W

= 3×10^‒3 Js^‒1

= 3×10^‒3×60 J min^‒1

The number of photons emitted per minute = 3×10^‒3×60 / 3.14×10^‒19

= 5.732 × 10¹⁷ photons/minute

ALITER

Intensity per unit area = No. of photons × Energy of a photon

No. of photons emitted per second (N) = Intensity per unit area / Energy of a photon

(or) No. of photons emitted per second (N) = Power output / Energy of a photon

N = 3×10^‒3 / 3.14×10^‒19

= 9.554 × 10¹⁵

No. of photons emitted per minute (N) = 9.554 × 10¹⁵ × 60

= 5.732 × 10¹⁷ photons/minute

The He‒Ne laser source emits 5.732 × 10¹⁷ photons per minute.

 

9. He‒Ne laser emits light at a wavelength of 632.8 nm and has an output power of laser is 5 mW. How many photons are emitted per second by this laser when operating?

Solution

Given Data:

Wavelength (λ) = 632.8 × 10^‒9 m

Power output = 5 × 10^‒3 Watts

Formula:

Energy of a photon E = hv = hc / λ

E = hc / λ

= 6.625×10^‒34×3×10⁸ / 632.8×10^‒9

E = 1.9875×10^‒25 / 632.8×10⁻⁹

E = 3.1408 × 10^‒19 Joules

∴ numbers of photons emitted per second

N = Power output / Energy of a photon

N = 5×10^‒3 / 3.1048×10^‒9

N = 1.5919 × 10^‒16   photons/second

Number of phtons emitted per second 1.5919 × 10⁻¹⁶

 

10. A Nd‒YAG laser emits light at wavelength of 1.063×10^‒6m If the output power is 20 W, then how many photons are emitted in ten minutes when the laser is in operation?

Given data

λ = 1.063 × 10^‒6 m

P = 20 W

Formula:

Number of photons =  ( Power output ) / ( Energy of a photon [hv] )

N = P / [hv / λ]

N = λP / hc

N = [ 1.063 × 10^‒6 × 20 ] / [ 6.625 × 10⁻³⁴ × 3 × 10⁸ ]

N = 2.126×10^‒5 / 1.9875×10^‒25

N = 1.0697×10²⁰ per second

∴ For 10 minutes N = 1.0697 × 10²⁰ × 60 × 10

∴ N = 6.418 × 10²²

∴ Number of photons emitted in 10 minutes = 6.418 × 10²²

 

11. Prove that laser action is not possible in optical frequencies under thermal equilibrium. (OR)

Show that the stimulated emission is not possible for:

(i) Sodium D line at 300° C and

(ii) At optical frequencies under thermal equilibrium.

Solution

Given data λ = 5000 Å; T = 300°K

We know R_sp/R_st = e ^{hv / KBT‒1}

= e^96.01 ‒ 1

R_sp/R_st = 4.9953 × 10⁴¹ ‒ 1

R_sp/R_st ≈ 4.9953 × 10⁴¹

Since the spontaneous emission is more predominant than stimulated emission, laser action is NOT POSSIBLE at optical frequencies under thermal equilibrium.

 

12. For a semiconductor laser, the bandgap is 0.9eV. What is the wavelength of light emitted from it. Use the following data:

Solution

C = 3 × 10⁸ m/s.

λ = 6.625 × 10^‒34 Js

Given Data

 E_g = 0.9 eV = 0.9 × 1.6×10^‒19 Joules

Formula:

Wavelength λ = hc / E_g = (6.625×10^‒34×3×10⁸) / (0.9×1.6×10^‒19)

 λ=1.3802×10^‒6m

The wavelength of light emitted = 1.3802 μm

 

13. For a heterojunction semiconductor laser, the band gap of the semiconductor used is 1.44 eV. By doping, the band gap of the semiconductor is increased by 0.2 eV. Calculate the change in the wavelength of the laser.

Solution

Given Data

 E_g1 = 1.44 eV = 1.44 × 1.6 × 10^‒19 J

 E_g1 = 2.309 × 10^‒19 J

E_g2 = 1.44 + 0.2

E_g2 = 1.64 eV = 1.64 × 1.6 × 10^‒19 J

E_g2 = 2.624 × 10^‒19 J

Formula:

We know

λ₁ = hc / E_g1

λ₂ = hc / E_g2

λ₁ ‒ λ₂ = hc [ 1/E_g1 ‒ 1/E_g2 ]

λ₁ ‒ λ₂ = 6.625 × 10^‒34 × 3 × 10⁸ × ( 1/2.309×10^‒19 ‒ 1/2.624×10^‒19 )

λ₁ ‒ λ₂ = 1.9875 × 10^‒25 × [4.33 × 10¹⁸ ‒ 3.81 × 10¹⁸]

λ₁ ‒ λ₂ = 1.987 × 10^‒25 × [5.2 × 10¹⁷] m

λ₁ ‒ λ₂ = 1033 × 10^‒10 m

λ₁ ‒ λ₂ = 1033Å

 Change in Wavelength = 1033Å.

 

14. For a laser at 2.0 m distance from the laser output beam spot diameter is 6.0 mm and beam, divergence is 1.2 mrad. Calculate the beam spot diameter at 5.0 m distance from the laser output.

Solution

Given Data

d₂ = 5m

d₁ = 2m

Diameter D₁ = 6 × 10⁻³ m

r₁ = 3 × 10^‒3 m

 ϕ = 1.2 mrad.

Formula

We know that ϕ = r₂‒r₁ / d₂‒d₁

ϕ = r₂‒r₁ / d₂‒d₁

r₂ = ϕ(d₂‒d₁) + r₁

r₂ = 1.2 × [5 – 2] + 3 × 10⁻³

r₂ =  3.603 m.

Diameter D₂ = 2 × 3.603

D₂ = 7.206m

Beam spot diameter at 5m = 7.206m.

 

 

ADDITIONAL SOLVED PROBLEMS

 

1. The first line of the principal series of Sodium D‒line at 5890Å. This corresponds to a transition from the first excited state to the ground state. What is the energy in electron volts of the first excited state?

Solution

The emitted energy E = hv = hc / λ

 = (6.625×10^‒34×3×10⁸ ) / (5890×10^‒10)

 = 3.3743 × 10^‒19 Joules

 = 3.3743×10^‒19 / 1.602×10^‒19

 = 2.1063 eV.

The Energy of the first excited state = 2.1063 eV.

 

2. What is the ratio of the stimulated emission to spontaneous emission at a temperature of 280°C for Sodium D‒line?

Solution

Stimulated emission / Spontaneous emission = R₂₁(ST) / R₂₁(SP)

 = 6.264 × 10^‒20

The ratio between the stimulated emission and spontaneous emission = 6.264 × 10^‒20

 

3. A CO₂ laser source emits light at a wavelength of 9.6 μm and has an output power of 10kW. How many photons are emitted in each hour by this laser while operating?

Solution

The frequency of the photon emitted by the laser beam = c/ λ

V = 3×10⁸ / 9.6×10^‒6

= 3.125 × 10¹³ Hz

We know that, The energy of a photon E = hv

(or) E = 6.625 × 10^‒34 × 3.125 × 10¹³

(or) E = 2.07 × 10^‒20 J

Energy emitted by the laser = 10 kilowatts.

= 10 × 10³ W = 10⁴ Js^‒1

= 10⁴ × 60 × 60 J hour^‒1

The number of photons emitted per hour = 10⁴×60×60 / 2.07×10^‒20

= 1.7391 × 10²⁷ photons per hour

The CO₂ laser source emits 1.739×10²⁷ photons per hour.

 

4. Examine the possibility of MASER action at thermal equilibrium using Einsteins theory of light.

Solution

We know

 е^hv/KBT ‒ 1 = R_sp / R_st

for microwave region λ = 10 cm ; T = 300° K

hv / K_BT = (6.625 × 10^‒34 × 3 × 10⁸) / (10×10^‒2×1.38×10^‒23×300)

R_sp / R_st = е^hv/KBT ‒ 1 = e^{(4.8 × 10‒4)} ‒ 1

= 1.00048 ‒ 1

R_sp / R_st = 4.8 × 10^‒4

Since the spontaneous emission is lesser than stimulated emission, MASER action is possible at thermal equilibrium.

 

5. An LED emits green light of wavelength λ = 5511.11Å. Find out the value of E_g.

Solution

E_g = hc / λ    Joules

E_g = [ hc / 1.602×10^‒19λ ] eV

E_g = [ 6.625×10^‒34×3×10⁸ / 1.602×10^‒19λ ] eV

E_g = 12406.36×10^‒10 / 5511.11×10^‒10          eV

E_g = 2.25 eV

 

ASSIGNMENT PROBLEMS

 

1. Calculate the ratio of the stimulated emission to spontaneous emission at a temperature 300°C for the Sodium D‒line. (Ans: 2.919×10^‒19)

 

2. Calculate how many photons are emitted in each minute in a He‒Ne laser source, which emits light at a wavelength of 6328Å. The output power of this source is 8mW. (Ans: 1.528×10¹⁸ photons/minute)

 

3. A ruby laser source emits light at a wavelength of 6943Å and has an output power of 10⁵ watts. How many photons are emitted in each hour by this laser while operating? (Ans: 1.257×10²⁷ photons per hour)

 

4. Transition occurs between a metastable state E₃ and an energy state E₂, just above the ground state. If emission is at 1.1 μm and E₂=0.4×10^‒19 J, find the energy of the E₃ state. (Ans: 2.2068×10^‒19 J)

 

5. If laser action occurs by the transition from an excited state to the ground state E₁ = 0, and if it produces light of 650 nm wavelength, what is the energy level of the excited state. (Ans: 3.0576×10^‒19 J)

 

6. Given E_g=1.43 eV. Find the wavelength of the light emitted by the diode for the given energy gap. Also find the colour of light emitted. (Ans: λ = 28671.32Å, IR)