In a nuclide, one $a.m.u.$ of mass is dissipated into energy to bind its nucleons. The energy equivalent of this mass is

  • A
    $931.5 \ eV$
  • B
    $931.5 \times 10^6 \ eV$
  • C
    $931.5 \times 10^6 \ MeV$
  • D
    $931.5 \ MV$

Explore More

Similar Questions

$^{14}C$ is

If ${}^{235}U$ is bombarded with neutrons,the atom will split into:

The atomic mass of an element is $12.00710 \, amu$. If there are $6$ neutrons in the nucleus of the atom of the element,the binding energy per nucleon of the nucleus will be $....... \, MeV$.
($e^{-} = 0.00055 \, amu$,$p = 1.00814 \, amu$,$n = 1.00893 \, amu$)

Difficult
View Solution

In the successive emission of $\alpha$- and $\beta$-particles,the number of $\alpha$- and $\beta$-particles that should be emitted for the conversion of ${}^{238}_{92}U$ to ${}^{206}_{82}Pb$ are:

Stable nuclides are those whose $n/p$ ratio is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo