Find the energy liberated in the reaction- 223Ra - 209Pb - 14CThe atomic masses needed are as follows-223-018u223Ra 208-981u209Pb 14-003u14C

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Standard XI

Physics

Nuclear Fission

Find the ener...

Question

Find the energy liberated in the reaction: $^{223} R a \to^{209} P b +^{14} C$
The atomic masses needed are as follows:
$^{223} R a 223.018 u$ $^{209} P b 208.981 u$ $^{14} C 14.003 u$

31.654 MeV

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32.654 MeV

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30.054 MeV

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None of these

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Solution

The correct option is A
31.654 MeV

Here, $Δ m = 223.018 - 208.981 - 14.003 = 0.034 u$
Hence, binding energy $= 0.034 \times 931 = 31.654 M e V$

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Similar questions

Q. Find the energy liberated in the reaction

^{223} R a \to^{209} P b +^{14} C

the atomic masses needed are as follows

^{223} R a^{209} P b^{14} C

22.018 u 208.981 u 14.003 u

Find the energy liberated in the reaction: $^{223} R a \to^{209} P b +^{14} C$
The atomic masses needed are as follows:
$^{223} R a 223.018 u$ $^{209} P b 208.981 u$ $^{14} C 14.003 u$

Q. A nuclear reaction along with the masses of the particles taking part in it is as follows:

A + B ⟶ C + D + Q M e V

1.002 a m u + 1.004 a m u ⟶ 1.001 a m u + 1.003 a m u + Q

The energy

Q

liberated in the reaction is

Q. Nuclear binding energy is the energy released during the hypothetical formation of the nucleus by the condensation of individual nucleons. Thus, binding energy per nucleon =

\frac{Total binding energy}{Number of nucleons}

For example, the mass of hydrogen atom is equal to the sum of the masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons, neutrons and electrons present. This difference in mass, termed as mass defect, is a measure of the binding energy of protons and neutrons in the nucleus. The mass-energy relationship postulated by Einstein is expressed as:

Δ E = Δ m c^{2}

Where

Δ E

is the energy liberated,

Δ m

is the loss of mass, and c is the speed of light.
In the reaction

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e +_{0}^{1} n

, if binding energies

_{1}^{2} H,_{1}^{3} H and_{2}^{4} H e

are respectively a, b and c (in MeV), then the energy released in this reaction is:

Q. Consider the fission reaction

{_{92} U}^{236} ⟶ X^{117} + Y^{117} + {_{0} n}^{1} + {_{0} n}^{1}

ie., two nuclei of same mass number

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are formed plus two neutrons. The binding energy per nucleon of

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8.5 M e V

where as of

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. Find the total energy liberated.

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Find the energy liberated in the reaction: 223Ra→209Pb+14C The atomic masses needed are as follows: 223Ra223.018u 209Pb208.981u 14C14.003u

The correct option is A 31.654 MeV Here, Δm=223.018−208.981−14.003=0.034 u Hence, binding energy =0.034×931=31.654 MeV

Find the energy liberated in the reaction: $^{223} R a \to^{209} P b +^{14} C$
The atomic masses needed are as follows:
$^{223} R a 223.018 u$ $^{209} P b 208.981 u$ $^{14} C 14.003 u$

The correct option is A
31.654 MeV

Here, $Δ m = 223.018 - 208.981 - 14.003 = 0.034 u$
Hence, binding energy $= 0.034 \times 931 = 31.654 M e V$