You visited us 1 times! Enjoying our articles? Unlock Full Access!

Binding Energy

The energy of...

Question

The energy of the reaction $L i^{7} + p ⟶_{2} H e^{4}$ is (the binding energy per nucleon in $L i^{7}$ and $H e^{4}$ nuclei are $5.60$ and $7.06 M e V$ respectively.)

17.3 M e V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.73 M e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.46 M e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D e p e n d s o n t h e b i n d i n g e n e r g y o f p r o t o n .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $17.3 M e V$
$\begin{matrix} Energy of r \times n & = B \cdot E_{p} - B \cdot E_{R} = 7.6 \times 4 \times 2 - 5.6 \times 7 = 17.3 M e v \end{matrix}$

Suggest Corrections

Similar questions

L i^{7} + p \to 2 H e^{4}

L i^{7}

H e^{4}

5.60 MeV

7.06 MeV

{Li}^{7} + p \to 2 {He}^{4}

{Li}^{7}

{He}^{4}

5.60 MeV

7.06 MeV

{L i}^{7}

{H e}^{4}

5.60 M e V

7.06 M e V

{L i}^{7} + P \to 2_{2} {H e}^{4}

If the binding energy per nucleon in Li⁷ and He⁴ nuclei are 5.60 MeV and 7.06 MeV , then what is the energy of the reaction ₃Li⁷ + ₁H¹ = 2 ₂He⁴ ?

_{7} 3 L i

5.60 M e V

7.06 M e V

p +_{3} 7 L i \to_{2} 4

Join BYJU'S Learning Program