The energy released when 7/17.13 kg of ⁷₃Li is converted into ⁴₂He by proton bombardment is α×10³² eV. The value of α is _ _. (Nearest integer) (Mass of ⁷₃Li=7.0183 u, mass of ⁴₂He=4.004 u, mass of proton =1.008 u and 1 u =931 MeV/c² and Avogadro number =6.0×10²³)

Atoms and Nuclei: JEE Main 2026 Physics question

JEE Main 2026 · 6 Apr, Morning shift · Q48

JEE Main 20266 Apr, Morning shiftPhysicsHardAtoms and Nuclei183× since 2002

The energy released when

\frac{7}{17.13}

kg of

\mathrm{^7_3Li}

is converted into

\mathrm{^4_2He}

by proton bombardment is

\alpha\times10^{32}

eV. The value of

\alpha

is ____. (Nearest integer) (Mass of

\mathrm{^7_3Li}=7.0183

u, mass of

\mathrm{^4_2He}=4.004

u, mass of proton

=1.008

u and 1 u

=931\ \mathrm{MeV/c^2}

and Avogadro number

=6.0\times10^{23}

)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2026 Atoms and Nuclei Question with Solution (6 Apr Morning shift, Q48) | JEE Arc