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Numerics on Lasers HRK-Ex-48-7 A three level laser emits light of wavelength 550 nm. (a) What is the ratio of population of the upper level (E2) to that of the lower level (E1) in laser transition, at 300 K? (b) At what temperature the ratio of the population of E2 to that of E1 becomes half? (c) At what negative temperature the population of the upper level exceeds that of the lower by 10%?

Solution N2 N1

(a) 

=

e



=

e

(E 2 -E1 ) kT

=e

 h c λkT

6.63×10-34 ×3×108 550×10-9 ×1.38×10-23 ×300

= 1.16×10

 38

Solution HRK-Ex-48-7 1(b)

 N2 = 0.5 = e N1 

i.e T =

e

h c λ k T

(E 2 -E1 ) k T

 0.5 ;

e



h c λ k T

h c  Ln (2) λ k T

h c = 37806K = 38000K λ k Ln (2)

HRK-Ex-48-7 1(C) N2  1.1  e N1 i.e

e

h c kT



( E2  E1 ) kT

 1.1 ;

e

h c kT

h c   Ln (1.1) kT

h c T   274945  280000 K  k Ln(1.1)

HRK-28 A ruby laser emits photons of wavelength 694.4nm. If the energy release per pulse is 150 mJ and lasts for 12.0 ps, calculate the (i) length of the pulse (ii) the number of photons in each pulse. Length of the pulse = Velocity  time or (duration) of the pulse = 3 108  1210-12 = 3.6 mm

No of photons/pulse 

N 

  

(Total energy/pulse)   hc   energy per photon       

(150  103 ) (6.63 1034 )(3 108 )   694.4  109 

 5.24  1017

HRK-30 A He-Ne laser emits light of wavelength of 632.8 nm and has an output power of 2.3 mW. How many photons are emitted each minute by this laser when operating? Energy = (power × time) = (2.3× 10-3 ) × 60 s

(Total energy per min) no of photons /min=  energy per photon  (2.3  10 3 )  60 N   (6.63  10 34 )(3  108 )    9 632.8  10    4.39  1017

HRK-31 A hypothetical atom has energy levels uniformly separated by 1 .2 eV. At a temperature of 2000 K, what is the ratio of the number of atoms in the 13th excited state to the number in the 11th excited state

N13  e N11 e

13.91

(E 2 -E1 )  kT

 ( 21.2)1.610 19

e

1.381023 2000

 9.098  107

HRK 33 A hypothetical atom has two energy levels with a transition wavelength of 582 nm. In such a sample at 300 K, 4 x 1020 atoms are there in the lower state. (a) How many occupy the upper state under conditions of thermal equilibrium? (b) Suppose, instead, that 3.0 x 1020 atoms are pumped into upper state, with 1.0 x 1020 remaining in the lower state. How much energy could be released in a single laser pulse?

Solution (a)

N2

 N1

e

h c kT

 (4  10 20 )   5.64  10

16

e

 (6.631034 )(3108 ) (58210 9 )(1.3810 23 ) k T

0

Solution

33(b)

Energy of each photon = hc



(6.63 1034 )(3 108 ) 19   3.42  10 J 9 (582 10 )

Total energy released

N

hc



 102 J

=

34 8 (6.63  10 )(3  10 )  (3  1020 ) (582 10 9 )

HRK -8-48 A laser emits at 424 nm in a single pulse that lasts 0.500 p,s. The power of the pulse is 2.80 MW. If we assume that the atoms contributing to the pulse underwent stimulated emission only once during the 0.500 s, how many atoms contributed?

Solution

HRK-8-52 (P-8) A hypothetical atom has only two atomic energy levels, separated by 3.2 eV. Suppose that at a certain altitude in the atmosphere of a star there are 6.1 x 1013/cm3 of these atoms in the higher-energy state and 2.5 x 1015/cm3 in the lower energy state. What is the temperature of the star's atmosphere at that altitude?

Solution

SJ-42.10-46 From the data shown in the figure, S.T the wavelength of the red HeNe laser light is approximately 633nm. 20.61eV

20.66eV 18.70eV

SJ 42-47 In CO2 laser the energy difference between the two levels is 0.117eV. Determine the frequency & the wavelength of the emitted radiation.

SJ 42-48: A Nd:YAG laser emits 3.00mJ pulse in 1.00ns, focused to a spot 30.0m in diameter on the retina. (i) Find the power irradiated per unit area at the retina. (ii) How much energy is delivered per pulse to an area of molecular size taken as a circular area 0.600nm in diameter? Ans : 4.24×1015 W/ m 2. and 1.2×10-12 J

(energy) (i) power per unit area =  Time  Area  3× 10-3 15 2   4.24× 10 W / m 1 109    (15  106 ) 2 (ii)energy delivered =(power  area  time) =4.24× 1015    (0.3  10 9 ) 2 110 9  1.2pJ

SJ 42-49 A ruby laser delivers a 10.0-ns pulse of 1.0MW average power. If the photons have a wavelength of 694.3nm, how many photons are contained in the pulse? Ans 3.49×1016.

SJ 61

A pulsed laser emits photons of wavelength 694.4nm. If the energy release per pulse is 3.00J and lasts for 14.0 ps, calculate the (a) length of the pulse (b) the number of photons in each pulse. (c) The beam has a circular cross section of diameter 0.600cm. Find the number of photons per cubic millimeter.

SJ 62 A pulsed laser emits photons of wavelength . If the energy release per pulse is E and lasts for t. Calculate the (a) length of the pulse (b) the number of photons in each pulse. (c) The beam has a circular cross section of diameter d. Find the number of photons per unit volume.

Compare the relative probabilities of spontaneous and stimulated emission in an equilibrium system at room temperature (T=300K) for transitions that occur in (a) the visible (h = 2eV) (b) the microwave regions (h = 104eV) of the spectrum. h   A21 kT   e  1  B21  ( )   2   A21 33 0.0259  e  1  10 B21  ( )  

Relative probability of the Spontaneous emission to = the Stimulated emission (a)

(b)

 A21  e B21  ( ) 

1104 0.0259

 3  1  3.8 10  

SPB The transition to the ground state from two closely spaced upper and lower states in a ruby laser results in the emission of photons of wavelengths 692.8nm and 694.3nm respectively. Calculate (a) the energy values of the two levels relative to the ground state and (b) also the ratio of their populations.

Energy released in transition from upper to ground state Upper

hc



(6.63  1034 )(3  108 )  (692.8  10 9 )(1.6  1019 )

Lower

 1.794eV  2.87  1019 J

694.3nm 692.8nm

Ground state

Energy in transition from lower to ground state is

h c



(6.63  10 34 )(3  108 )  (694.3  109 )(1.6  1019 )

 1.788 eV  2.865  1019 J

Ratio of population between two levels

N2 e N1



( E2  E1 ) kT

e



(1.794 1.79) 0.0259

 0.8568

SPB A ruby laser delivers a 20.0-ns pulse of 0.1MW average power/pulse. If the number of photons emitted per second is 6.98×1015. Calculate wavelength of the photons. Energy =power ×time Energy per photon = (Energy released per pulse ÷ no of photons per pulse)

(0.1 106 )  (20 10 9 )  (6.98  1015 )  2.865  10 

19

J;

i.e

hc



 2.865  10 19 J

(6.63  10 34 )(3  108 )   694.3nm 19 2.865  10

A pulsed laser emits photons of wavelength 780 nm with 20MW average power per pulse. Calculate the number of photons in each pulse if the pulse duration is 10ns. (Dec 08-odd)

(Total energy/pulse) No of photons/pulse    hc   energy per photon        N

  

(20 106 109 ) (6.63 1034 )(3 108 )   780 109 

 7.84 1016