Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the energy released when three $\alpha$-particles combined to form a ${ }^{12} \mathrm{C}$ nucleus, the mass defect is
(Atomic mass of ${ }_{2} \mathrm{He}^{4}$ is $4.002603 \mathrm{u}$ )
Options:
(Atomic mass of ${ }_{2} \mathrm{He}^{4}$ is $4.002603 \mathrm{u}$ )
Solution:
1088 Upvotes
Verified Answer
The correct answer is:
$4.002603 \mathrm{u}$
Mass defect
$$
\begin{array}{l}
\begin{array}{l}
\Delta \mathrm{m}=\text { Total mass of } \alpha \text {-particles } \\
\quad-\text { mass of }^{12} \mathrm{C} \text { nucleus } \\
=3 \times 4.002603-12=12.007809-12 \\
=0.007809 \text { unit }
\end{array}
\end{array}
$$
$$
\begin{array}{l}
\begin{array}{l}
\Delta \mathrm{m}=\text { Total mass of } \alpha \text {-particles } \\
\quad-\text { mass of }^{12} \mathrm{C} \text { nucleus } \\
=3 \times 4.002603-12=12.007809-12 \\
=0.007809 \text { unit }
\end{array}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.