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scoray [572]
2 years ago
14

The mass of a proton is 1.00728 amu and that of a neutron is 1.00867 amu. what is the binding energy (in j) of a 61co nucleus? (

the mass of a cobalt-61 nucleus is 60.9325amu.)
Chemistry
1 answer:
bazaltina [42]2 years ago
5 0

With the given equations the binding energy is 1.372 *10^{-12} J.

<h3>How will you calculate binding energy using the given equations?</h3>

Step 1: The given data is:

The proton's mass 1.00728 amu

The neutron's mass 1.00867 amu

The mass of a cobalt-61 nucleus is 60.9325 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt61 has 27 protons and 34 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 34 neutrons = 34*1.00867 u = 34.29478 u

Total mass of protons + neutrons = 27.19656 u + 34.29478 u = 61.49134 u

Mass of a cobalt61 nucleus = 60.9325 amu

Mass defect = Δm = 0.55884 u

ΔE =c²*Δm

ΔE = (3.00 *10^8 m/s^2) *(0.55884 amu))*(1.00 g/ 6.02 *10^{23} amu)*(1kg/1000g)=8.36 * 10^{-11} J

Step 3: Calculate binding energy per nucleon

ΔE = 8.36* 10^{-11} J

binding energy= \frac{8.36* 10^{-11}}{60.9325}  J = 1.372 *10^{-12} J

The binding energy per nucleon = 1.372 *10^{-12} J

To learn more about binding energy calculation, click here:

brainly.com/question/14019034

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Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

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