Home
Class 12
CHEMISTRY
In the nuclear fission .(1)H^(2) + .(1)H...

In the nuclear fission `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(4)` the masses of `._(1)H^(2) and ._(2)he^(4)` are 2.014 mu and 4.003 mu respectively. The energy released/atom of helium formed is ....MeV

A

16.76

B

26.38

C

13.26

D

23.275

Text Solution

Verified by Experts

The correct Answer is:
D
`Deltam = (2 xx 2.014- 4.003) "mu" = 0.025 `mu
1 amu = 931.5 MeV
`E= 0.025 xx 931.5`
E per nucleon `= (931.5)/(4) = 23.4 MeV ` or 23.28
Promotional Banner

Similar Questions

Explore conceptually related problems

._(1)^(3)H and ._(2)^(4)He are:

The nuclear reaction ._(1)^(2)H + ._(1)^(2)H to ._(2)^(4)He is called

The binding energy per nucleon for ""_(1)H^(2) and ""_(2)He^(4) are 1.1 MeV and 7.1 MeV respectively. The energy released when two ""_(1)H^(2) to form ""_(2)He^(4) is ………….. MeV.

Calculate the energy released in MeV during the reaction ._(3)^(7)Li + ._(1)^(1)H to 2[._(2)^(4)He] if the masses of ._(3)^(7)Li, ._(1)^(1)H and ._(2)H_(4)He are 7.018, 1.008 and 4.004 amu respectively.

Calcualte the energy released (in joule and MeV ) in the follwing nulcear reaction: ._(1)^(2)H + ._(1)^(2)H rarr ._(2)^(3)He + ._(0)^(1)n Assume that the masses of ._(1)^(2)H, ._(2)^(3)He and neutron (n) are 2.0141,3.0160 and 1.0087 respectively in amu.

Calculate the energy released in joules and MeV in the following nuclear reaction: ._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1) Assume that the masses of ._(1)H^(2) , ._(2)He^(3) , and neutron (n) , respectively, are 2.40, 3.0160, and 1.0087 in amu.

In the nuclear raction given by ._2He^4 +._7N^(14) rarr ._1H^1 +X the nucleus X is