Question

What would be the energy required to dissociate completely $1$ g of $Ca\left(40\right)$ into its constituent particles?
Given: Mass of proton = 1.007277 amu,
Mass of neutron = 1.00866 amu,
Mass of Ca-40= 39.97545 amu
(take amu = 931 MeV)

• A. $4.813\times {10}^{24}$MeV
• B. $4.813\times {10}^{24}$ eV
• C. $4.813\times {10}^{23}$
• D. None of the above

Answer 1

The correct option is A $4.813\times {10}^{24}$MeV

Given,

$m_p=1.007277amu$ mass of proton

$m_n=1.00866amu$ mass of neutron

$m_c(ca-40)=39.97545amu$

and $1amu=931MeV/c^2$

$Z=20$ No of proton for calcium

$N=20$ no. of neutron

The energy required to dissociate completely is called binding energy.

$B.E=(Z{m}_p+N{m}_n-m_c)c^2$ . . . .(1)

$B.E=(20\times 1.007277+20\times 1.00866-39.97545)c^2$

$B.E=0.34359\times 931MeV$

$B.E=319.6MeV$ This is for 1 atom of calcium-40

No. of atom in $1g$ of Ca-40 is,

$N=\frac{1}{40}\times 6.6022\times {10}^{23}$

$N=1.65055\times {10}^{22}$ atom.

The total energy required is,

$E=N(B.E)$

$E=1.65055\times {10}^{22}\times 319.6MeV$

$E=4.813\times {10}^{24}MeV$

The correct option is A.