UN Fisika 2016 Paket 3 Nomor 39

Perhatikan reaksi fusi berikut!

{}_{2}^{3}He+{}_{2}^{3}He\to ~2{}_{1}^{1}H+x+E

Diketahui:

{}_{1}^{1}H=1,0081~sma

{}_{2}^{4}He=4,0039~sma

{}_{2}^{3}He=3,0158~sma

Jika 1 sma = 931 MeV, energi yang dilepaskan pada reaksi fusi tersebut adalah …

A.    1887,7887 MeV D.    19,9211 MeV
B.     967,8676 MeV E.     10,7065 MeV
C.     949,2476 MeV

Penyelesaian

Energi ikat inti.

E=\text{ }\!\!\Delta\!\!\text{ }m.{{c}^{2}}

E=\text{ }\!\!\Delta\!\!\text{ }m.~931~MeV

E=\left\{ 2\left( 3,0158 \right)-(2\left( 1,0081 \right)+4,0039) \right\}\times 931~MeV

E=10,7065~MeV

jawaban E

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