1000 Solved Problems in Modern Physics

414 7 Nuclear Physics – I

T 2 =hc/λ 2 −W

T 2 −T 1 =hc(1/λ 2 − 1 /λ 1 )

(4. 0 − 1 .8)× 1. 6 × 10 −^19 J=hc

(

1

700 × 10 −^10 −

1

800 × 10 −^10

)

Solving forh, we geth= 6. 57 × 10 −^34 Js

The accepted value ish= 6. 625 × 10 −^34 J-s

7.73hν 0 =W= 4 .7eV
λ 0 = 1 , 241 / 4. 7 =264 nm.

7.74 TA= 4. 25 −WA (1)

TB= 4. 7 −WB (2)

TA

TB

=

PA^2

PB^2

=

λ^2 B

λ^2 A

=4(3)

TB=TA− 1 .5(4)

Solving the above equations,

TA=2 eV andTB= 0 .5eV

WA= 2 .25 eV andWB= 4 .2eV

7.3.6 Pair Production ....................................

7.75 The minimum photon energy required for the e−e+pair production is 2mc,
wheremis the mass of electron. Therefore
hν= 2 mc^2 = 2 × 0. 511 = 1 .022 MeV
The corresponding wavelength
λ=(1, 241 / 1. 022 × 106 eV) nm= 1. 214 × 10 −^12 m.

7.76 In the annihilation process energy released is equal to the sum of rest mass
energy of positron and electron which is 2mc^2. Because of momentum and
energy conservation, the two photons must carry equal energy. Therefore each
gamma ray carries energy
Eγ=mc^2 = 0 .511 MeV
The wavelength of each photon
λ= 1 , 241 /(0. 511 × 106 )nm= 2. 428 × 10 −^3 nm= 2. 428 × 10 −^12 m

7.77 Let us suppose that the e+e−pair is produced by an isolated photon of energy
E=hνand momentumhν/c. Let the electron and positron be emitted with
momentump−andp+, their total energy beingE−andE+. Energy conserva-
tion gives
hν 0 =E++E− (1)
The momentum conservation implies that the three momenta vectors,p 0 ,p+
andp−must form the sides of a closed triangle, as in Fig. 7.17. Now in any
triangle, any side is equal or smaller than the sum of the other sides. Thus