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Fisika Inti

1. This document provides calculations of Q values, kinetic energies, and velocities for various nuclear decay processes, including: - Bk-247 decaying to Cm-243 and an alpha particle - Cf-251 decaying to Cm-247 and an alpha particle - Th-230 decaying to Ra-226 and an alpha particle 2. It also calculates the mass of Cm-242 given the maximum alpha energy observed in its decay to Pu-238, and the mass of Th-234 given the maximum alpha energy observed in the decay of U-238 to Th-234.

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Miftah Nur Wulan
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296 views3 pages

Fisika Inti

1. This document provides calculations of Q values, kinetic energies, and velocities for various nuclear decay processes, including: - Bk-247 decaying to Cm-243 and an alpha particle - Cf-251 decaying to Cm-247 and an alpha particle - Th-230 decaying to Ra-226 and an alpha particle 2. It also calculates the mass of Cm-242 given the maximum alpha energy observed in its decay to Pu-238, and the mass of Th-234 given the maximum alpha energy observed in the decay of U-238 to Th-234.

Uploaded by

Miftah Nur Wulan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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TUGAS FISIKA INTI

Oleh : Samia Siti Marsiah (1700937)

1. Tentukan nilai Q pada peluruhan berikut:


a. Bk 247 →C +α
Jawab:
Q=¿
Q=[ ( 247,070300 )−( 243,061375 )− ( 4,002603 ) ]
Q=0,006322C 2 × 931,494 MeV /C 2
Q=5,88890507 MeV

b. Cf 251 →Cm 247 +α


Jawab:
Q=¿
Q=[ ( 251,079580 )−( 247,070347 ) −( 4,002603 ) ]
Q=0,006603C 2 × 931,494 MeV /C 2
Q=6,17580522 MeV

c. Th230 → Ra226 +α
Jawab:
Q=¿
Q=[ ( 230,033128 )−( 226,025403 )− ( 4,002603 ) ]
Q=0,005122C 2 × 931,494 MeV /C 2
Q=4,77111227 MeV

2. For each decay given in problem1, calculate the kinetic energy and velocity of the
daughter nucleus after the decay.
A−4 247−4
a. k α = Q= × 5,88890507=5,79353818 MeV
A 247
Keceptannya
2K 2(5,79353818)
v=
√ m( Am)
=
√243,061375 ×931,494
=√ 5,11773577 ×10−5=0,0071538 c

A−4 247−4
b. k α = Q= × 6,17580522=6,07738602 MeV
A 247
Keceptannya
2K 2(6,077386028)
v=
√ m( Am)
=
√
247,070347 × 931,494
=√ 5,28136439× 10−5=0,00726729c

A−4 247−4
3. k α = Q= × 4,77111227=4,68813639 MeV
A 247
Keceptannya
2K 2(4,688136398)
v=
√ m( Am)
=
√
226,025403 ×931,494
=√ 4,45370117 × 10−6=0,00211037939 c

4. From the known atomic masses, compute the Q values of the decays
a. Pu242 →U 238 +α
Jawab:
Q=¿
Q=[ ( 242,058738 )−( 238,050785 )− ( 4,002603 ) ]
Q=0,00535C 2 × 931,494 MeV /C 2
Q=4,9834929 MeV
b. Po208 → Pb 204 + α
Jawab:
Q=¿
Q=[ ( 207,981222 )−( 203,973020 )−( 4,002603 ) ]
Q=0,005599C 2 × 931,494 MeV /C 2
Q=5,21543491 MeV
c. Po208 → Pt 196 +C 12
Jawab:
Q=¿
Q=[ ( 207,981222 )−( 195,964926 )−( 12,000000 ) ]
Q=0,016296C 2 × 931,494 MeV /C 2
Q=415,1796262 MeV
d. Bi210 → Pb208 + H 2
Jawab:
Q=¿
Q=[ ( 209,984095 )−( 207,976627 ) −( 2,014102 )]
Q=−0,006634 C2 ×931,494 MeV / C2
Q=−6,1795321 MeV
5. In the decay of C m242 to Pu238 , the maximum α energy is 6112,9 ± 0,1 keV . Given the

Maa of Pu238, find the mass of C m242.


Jawab:
Q=( m x −m x −mα ) 931,494 MeV / u
'

( 6112,9 ± 0,1 ) 10−3 MeV =( Cm242 −238,049555−4,002603 ) u .931,494 MeV /u


( 6112,9 ± 0,1 ) 10−3 MeV =931.404 Cm242 −221741,73218517−3728,400678
( 6112,9 ± 0,1 ) 10−3 MeV =931.494 Cm242 −225569,400678
C m242=242,05793436± 0,1 u
6. The highest energy α particle emitted in the decay of U 238 to T h234 is 4196 ± 4 keV .

from this information and the known mass of U 238, compute the mass of T h234.
Jawab:
Q=( m x −m x −mα ) 931,494 MeV / u
'

( 4196 ± 4 ) 10−3 MeV =( T h234 −238,050785−4,002603 ) u .931,494 MeV /u


( 4196 ± 4 ) 10−3 MeV =931.494 T h234 −221742,87792279−3728,400678
( 4196 ± 4 ) 10−3 MeV =931.494 T h234 −225471,27860079
T h234 =234,043677 ± 4 u

7.

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