Which of the following nuclei would be the least stable a 2 protons, 2 neutrons b 1 proton 1 neutron c 1 proton 3 neutrons d 1 proton 2 neutrons

C. 1 proton 3 neutrons

Explanation:

A nucleus is more stable if the ratio of the neutrons to protons is between 1:1 and 1:1.5.

Thus the ratios of neutrons to protons for  the nuclei are as follows

A- 1:1

B- 1:1

C- 1:3

D- 1:2

Among these ratios, C is the greatest thus the nucleus is the least stable.

The nuclei with D. 1 proton and 2 neutrons will be least stable.

Nuclides containing odd quantities of the two protons and neutrons are the least steady and this implies increasingly radioactive.  

Nuclides containing even quantities of the two protons and neutrons are most steady and this implies less radioactive.  

Nuclides contain odd quantities of protons (positive charged particles) and even quantities of neutrons are less steady than nuclides which are containing even quantities of protons and odd quantities of neutrons. By and large, atomic strength is more prominent for nuclides containing even quantities of protons and neutrons or both.

When we add 1 proton with 2 neutrons we get 3 as answer, which means an odd value, thus nuclide with this proportion of proton and neutron is least stable.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

2b2t

explanation:

2b2t