You visited us 1 times! Enjoying our articles? Unlock Full Access!

Binding Energy

The binding e...

Question

The binding energy of $_{17}^{35} C l$ nucleus is 298 MeV. Find its atomic mass. Given, mass of a proton $(m_{p}) = 1.007825 a m u$ , mass of neutron $(m_{n}) = 1.008665 a m u$

Open in App

Solution

Here proton number $Z = 17$ and neutron number, $N = A - Z = 35 - 17 = 18$
Binding energy, $B E = Δ m c^{2}$
Mass defect , $Δ m = 298 M e V / c^{2} = 298 / 931.5 a m u = 0.3199 a m u$ as $(1 a m u = 931.5 M e V / c^{2})$
If M be atomic mass, $Δ m = (17 m_{p} + 18 m_{n}) - M$
or $M = (17 \times 1.007825 + 18 \times 1.008665) - 0.3199 = 34.9597 a m u$

Suggest Corrections

Similar questions

_{17} {C l}^{35}

298

(_{1} H^{1})

1.008143

1

= 931

_{17} {C I}^{35}

298 M e V .

({_{1} H}^{1})

1.008143 a m u

1.008986 a m u .

1 a m u = 931 M e V .

M_{p}

M_{n}

B

Z

N

M (N, Z)

(c is the velocity of light)

_{50}^{120} S n

_{50}^{120} S n =

119.902199 a m u

= 1.007825 a m u

= 1.008665 a m u

_{50}^{120} S n

m_{p} = 1.00783 U

m_{n} = 1.00867 U

m_{S n} = 119.902199 U

1 U = 931 MeV)

Join BYJU'S Learning Program