Question

[Hoàng Đức Long]

Cho phản ứng hạt nhân ${}_{17}{}^{31}\mathrm{Cl}+{}_1{}^1\mathrm{p}\to {}_0{}^1\mathrm{n}+{}_{18}{}^{37}\mathrm{Ar}$. Biết ${\mathrm{m}}_{\mathrm{Ar}}=36,956889\left(\mathrm{u}\right)$,  ${\mathrm{m}}_{\mathrm{Cl}}=36,956563\left(\mathrm{u}\right)$,  ${\mathrm{m}}_{\mathrm{P}}=1,007276\left(\mathrm{u}\right)$,  ${\mathrm{m}}_{\mathrm{n}}=1,008665\left(\mathrm{u}\right)$, $1\mathrm{u}=1,6605.{10}^{-27}\left(\mathrm{kg}\right)$, $\mathrm{c}=3.{10}^8\mathrm{m}/\mathrm{s}$. Phản ứng trên tỏa hay thu năng lượng, bao nhiêu Jun(J)?

A. Thu  $2,56.{10}^{-13}\left(\mathrm{J}\right)$

B. Thu  $-2,56.{10}^{-13}\left(\mathrm{J}\right)$

C. Thu  $8,5.{10}^{-22}\left(\mathrm{J}\right)$

D. Thu  $-8,5.{10}^{-22}\left(\mathrm{J}\right)$

Answer 1

[Vũ Thành Nam]

Chọn B.