Question

If $x$ gm of a nuclear fuel of mass number $A$ undergoes fission inside a reactor, then the number of fission will be ($N$-Avogadro number)

• A. $\frac{NA}{x}$
• B. $NAx$
• C. $\frac{Nx}{A}$
• D. $\frac{Ax}{N}$

Answer 1

The correct option is C $\frac{Nx}{A}$

Power of a nuclear reactor :
The output power of a nuclear reactor is given by:
$P=nE$ ----(1)
Where, $n=$ No. of fission taking place in 1 sec
$E=$ Energy released per fission.
If ${}^{\prime }{x}^{\prime }$ grams of a nuclear fuel of mass number 'A' undergo fission in a time of ${}^{\prime }{t}^{\prime }$ sec and${}^{\prime }{E}^{\prime }$ is energy released per fission, then the power output of the nuclear reactor is given by:

$P=\frac{NEx}{At}$ --(2) where N - Avagadro number.
Comparing equation (1) and (2) we get:

$n=\frac{Nx}{At}$
Where $n$ is no. of fission taking place in $1$ sec.
So, for '$t$' sec time, no. of fission on '$t$' sec will be
$\frac{Nx}{At}t=\frac{Nx}{A}$

So, correct choice is option C.