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Question Number 154554 by peter frank last updated on 19/Sep/21

Commented by peter frank last updated on 19/Sep/21

$_{2}^{3} He = 3 . 01664 U$
	Answered by peter frank last updated on 19/Sep/21

	$Calculate the binding energy$ $per nuclei for the_{2}^{4} He and_{2}^{3} He$
	Answered by peter frank last updated on 20/Sep/21

	$_{1}^{1} H = 1 . 007834 U$ $_{2}^{3} He = 3 . 01664 U$ $_{2}^{4} He = 4 . 00387 U$ $_{0}^{1} η = 1 . 008674 U$ $_{2}^{4} He \to 2_{1}^{1} H + 2_{0}^{1} η$ $_{2}^{4} He (mass) = m_{1}$ $2_{1}^{1} H + 2_{0}^{1} η (mass) = m_{2}$ $△ m = m_{1} - m_{2} = 0 . 029134 U$ $from$ $1 U = 931 Mev$ $0 . 029134 U = ?$ $Binding energ = 931 \times 0 . 029134 U = 27 . 12 Mev$ $Binding energy per nucleon$ $= \frac{27.12}{4} = 6 . 78 Mev / nucleon$ $\underset{m_{1}}{\underset{⏟}{_{2}^{3} He}} \to \underset{m_{2}}{\underset{⏟}{2_{1}^{1} H + 2_{0}^{1} η}}$ $△ m = m_{1} - m_{2} = 7.69 \times 10^{- 3} U$ $1 U = 931 Mev$ $7.69 \times 10^{- 3} U = ?$ $B . E = 7 . 15939 Mev$ $B . E / Nucleon = \frac{7 . 15939 Mev}{3} = 2 . 386 Mev / nucleon$
	Commented by Tawa11 last updated on 21/Sep/21

	$nice sir$
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