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  4. The threshold energy (in MeV) for the following nuclear reaction to proceed is 24 He + 714 N arrow 817 O + 11 H Atomic mass of 24 He =4.00260 amu Atomic mass of 714 N =14.00307 amu Atomic mass of 817 O =16.99913 amu Atomic mass of 11 H =1.00783 amu

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Q. The threshold energy (in MeV) for the following nuclear reaction to proceed is
${ }_{2}^{4} He +{ }_{7}^{14} N \rightarrow{ }_{8}^{17} O +{ }_{1}^{1} H$
Atomic mass of ${ }_{2}^{4} He =4.00260\, amu$
Atomic mass of ${ }_{7}^{14} N =14.00307\, amu$
Atomic mass of ${ }_{8}^{17} O =16.99913\, amu$
Atomic mass of ${ }_{1}^{1} H =1.00783\, amu$

Nuclei

Solution:

$E_{\text {thereshold }}=(-Q)+(-Q)\left(\frac{m_{ He }}{m_{ N _{2}}}\right)$
$Q=\left(m_{ He }+m_{ N _{2}}-m_{ O }-m_{ H }\right) c^{2}$