Question

The overall reaction taking place during hydrogen burning in a young star is

$4{ }_1^{1}H{\to }_2^{4}He+2{ }_1^0\beta +2 \nu +energy$

How much energy (in MeV) is released per He nucleus formed? Per mole of He? (Masses: ₁¹H atom=1.007825 amu ; ⁴₂He atom-4.00260 amu ; positron $-5.48580\times {10}^{-4}amu$.)

Short answer

The energy (in MeV) is released per He nucleus formed is 25.71 MeV and the energy released per mole of helium is $1.548\times {10}^{25}MeV$.

Step-by-step solution

Definition of Concept

Nuclear reaction: A nuclear reaction occurs when a lighter nucleus fuses together to form new stable nuclei, or when a heavier nucleus splits into stable daughter nuclei, releasing a large amount of energy.

Calculating the mass difference

Considering the given information:

$4_1^{1}H{\to }_2^{4}He+2_1^0\beta +2v+energy$

$\Delta m$is the difference in the masses of the reactants and products..

$$\begin{gathered} 4_1^{1}H{\to }_2^{4}He+2_1^0\beta +2v+energy \\ {m}_{reac\tan t}= 4\times m_H=4\times 1.007825amu \\ = 4.0313amu \\ {m}_{product}=m_{He}+2\times m_e \\ =4.00260amu+2\times 5.4858\times {10}^{-4}amu \\ =4.0037amu \\ \Delta m=m_{rcac\tan t}-m_{product} \\ =4.0313amu-4.0037amu \\ =0.0276amu \end{gathered}$$