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Obtain the Maximum Kinetic Energy Of β-particles, and the Radiation Frequencies Of γDecays in the Decay Scheme Shown in Fig. 13.6. You Are Given that - Physics

Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γdecays in the decay scheme shown in Fig. 13.6. You are given that

(198Au) = 197.968233 u

(198Hg) =197.966760 u

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Solution

It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ1-decay is given as:

E1 = 1.088 − 0 = 1.088 MeV

hν1= 1.088 × 1.6 × 10−19 × 106 J

Where,

h = Planck’s constant = 6.6 × 10−34 Js

ν= Frequency of radiation radiated by γ1-decay

`:. v_1 = E_1/h`

`= (1.088 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 2.637 xx 10^20 Hz`

It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ2-decay is given as:

E2 = 0.412 − 0 = 0.412 MeV

hν2= 0.412 × 1.6 × 10−19 × 106 J

Where,

ν= Frequency of radiation radiated by γ2-decay

`:. v_2 = E_2/h`

`= (0.412 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 9.988 xx 10^(19) Hz`

It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to γ3-decay is given as:

E3 = 1.088 − 0.412 = 0.676 MeV

hν3= 0.676 × 10−19 × 106 J

Where,

ν= Frequency of radiation radiated by γ3-decay

`:. v_3 =  E_3/h`

`= (0.676 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 1.639 xx 10^20 Hz`

Mass of `m(""_78^198Au)`= 197.968233 u

Mass of `m(""_80^198 Hg)`= 197.966760 u

1 u = 931.5 MeV/c2

Energy of the highest level is given as:

`E = [m(""_78^198 Au) - m(""_80^190 Hg)]`

`= 197.968233 - 197.966760 = 0.001473 u`

`= 0.001473 xx 931.5 = 1.3720955 MeV`

βdecays from the 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the βparticle = 1.3720995 − 1.088

= 0.2840995 MeV

βdecays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the βparticle = 1.3720995 − 0.412

= 0.9600995 MeV

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Q 29 | Page 465
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