JEE PHYSICS MCQ: NUCLEI

1 Medium

1 of 1

Application +4 | -1

Calculate the disintegration energy Q for the fission of

{ ~}_{42}^{98} \mathrm{M}

o into two equal fragments,

{ }_{21}^{49} \mathrm{~S} c

. If

Q

turns out to be positive, explain why this process does not occur spontaneously. Given that :

m\left({ }_{42}^{98} \mathrm{Mo}\right) =97.90541 \mathrm{~amu}

m\left({ }_{21}^{49} \mathrm{Sc}\right) =48.95002 \mathrm{~amu}

m_n =1.00867 \mathrm{~amu}

View answer and explanation

Correct option: B

944.6 \mathrm{~MeV}

Option B (

944.6 \text{{ MeV}}

) is correct because it is the disintegration energy Q calculated using the mass defect and the conversion factor from amu to MeV.

More Questions from NUCLEI