Chapter 19, Problem 1SP

(a)

To determine

The neutron and proton numbers for carbon (C), nitrogen (N), and oxygen (O).

Answer to Problem 1SP

The neutron number for carbon (C) is 6 and proton number of carbon is 6.

The neutron number for Nitrogen (N) is 6 and proton number of Nitrogen is 7.

The neutron number for Oxygen (O) is 6 and proton number of Oxygen is 8.

Explanation of Solution

The number of proton number is equal to the atomic number and proton number plus neutron number is equal to atomic mass. Therefore mass number minus atomic number will give neutron number.

From the periodic table, the proton number of carbon (C) is 6 and mass number is 12.

Therefore neutron number is $12-6=6$.

The proton number of Nitrogen (N) is 7 and mass number is 14.

Therefore neutron number is $14-7=7$.

The proton number of (O) Oxygen is 8 and mass number is 16.

Therefore neutron number is $16-8=8$.

Thus, the neutron number for carbon (C) is 6 and proton number of carbon is 6.

The neutron number for Nitrogen (N) is 6 and proton number of Nitrogen is 7.

The neutron number for Oxygen (O) is 6 and proton number of Oxygen is 8.

(b)

To determine

The ratio of neutrons and protons for the stable isotopes of carbon, Nitrogen and Oxygen.

Answer to Problem 1SP

The ratio of neutrons and protons for the stable isotopes of carbon, Nitrogen and Oxygen are $1.0$ for each.

Explanation of Solution

The stable isotopes of carbon, Nitrogen and Oxygen are carbon-12, Nitrogen-14 and Oxygen-16 respectively.

In the case of carbon-12 , the neutron number is 6 and proton number is $6$.

Write the expression for the ratio of neutron to proton of isotope.

$\text{Ratio}\text{=}\frac{N_n}{N_p}$ . (1)

Here,

$N_n$ is the number of neutrons

$N_p$ is the number of protons

Substitute $6$ for $N_n$ and $6$ for $N_p$ in the above equation to get $\text{ratio of neutron to proton}$.

$\begin{array}{c}\text{Ratio}\text{=}\frac{6}{6}\\ =1.0\end{array}$

In the case of Nitrogen-14, the neutron number is $7$ and proton number is $7$.

Substitute $7$ for $N_n$ and $7$ for $N_p$ in equation (1) to get ratio of neutron to proton for Nitrogen.

$\begin{array}{c}\text{Ratio}=\frac{7}{7}\\ =1.0\end{array}$

In the case of Oxygen-16, the neutron number is $8$ and proton number is $8$.

Substitute $8$ for $N_n$ and $8$ for $N_p$ in the above equation to get ratio of neutron to proton for Oxygen.

$\begin{array}{c}\text{Ratio}=\frac{8}{8}\\ =1.0\end{array}$

Conclusion:

Thus, the ratio of neutrons and protons for the stable isotopes of carbon-12, Nitrogen-14 and Oxygen-16 are $1.0$ for each.

(c)

To determine

The neutron and proton numbers for silver (Ag), cadmium (Cd), and indium (In).

Answer to Problem 1SP

The neutron number for silver (Ag) is 61 and proton number of silver is 41.

The neutron number for cadmium (Cd) is 64 and proton number of cadmium is 48.

The neutron number for indium (In) is 66 and proton number of indium is 49.

Explanation of Solution

The number of proton number is equal to the atomic number and proton number plus neutron number is equal to atomic mass. Therefore mass number minus atomic number will give neutron number.

From the periodic table, Atomic number of silver is $47$ and mass number is $108$.

Therefore number proton number is $47$.

Therefore neutron number is $108-47=61$.

From the periodic table, Atomic number of cadmium is $48$ and mass number is $112$.

Therefore number proton number is $48$.

Therefore neutron number is $112-48=64$.

From the periodic table, Atomic number of indium is $49$ and mass number is $115$.

Therefore number proton number is $49$.

Therefore neutron number is $115-49=66$.

Thus, the neutron number for silver (Ag) is 61 and proton number of silver is 41.

The neutron number for cadmium (Cd) is 64 and proton number of cadmium is 48.

The neutron number for indium (In) is 66 and proton number of indium is 49.

(d)

To determine

The ratio of neutrons and protons for the stable isotopes of silver, cadmium and indium.

Answer to Problem 1SP

The ratio of neutrons and protons for the stable isotope of silver is $1.30$.

The ratio of neutrons and protons for the stable isotope of cadmium is $1.33$.

The ratio of neutrons and protons for the stable isotope of indium is $1.35$.

The average value of ratio is $1.33$.

Explanation of Solution

The stable isotopes of silver, cadmium and indium are $\text{silver-108}$, $\text{cadmium-112}$ and $\text{indium-115}$ respectively.

In the case of $\text{silver-108}$, the neutron number is 61 and proton number is $47$.

Write the expression for the ratio of neutron to proton of isotope.

$\text{Ratio}\text{=}\frac{N_n}{N_p}$ . (1)

Substitute $61$ for $N_n$ and $47$ for $N_p$ in the above equation to get $\text{ratio of neutron to proton}$.

$\begin{array}{c}\text{Ratio}\text{=}\frac{61}{47}\\ =1.30\end{array}$

In the case of $\text{cadmium-112}$, the neutron number is $64$ and proton number is $48$.

Substitute $64$ for $N_n$ and $48$ for $N_p$ in equation (1) to get ratio of neutron to proton for $\text{cadmium-112}$.

$\begin{array}{c}\text{Ratio}=\frac{64}{48}\\ =1.33\end{array}$

In the case of $\text{indium-115}$, the neutron number is $66$ and proton number is $49$.

Substitute $8$ for $N_n$ and $8$ for $N_p$ in the above equation to get ratio of neutron to proton for $\text{indium-115}$.

$\begin{array}{c}\text{Ratio}=\frac{66}{49}\\ =1.35\end{array}$

Write the average value of ratios.

$\begin{array}{c}\text{average}\text{=}\frac{\text{1}\text{.30+1}\text{.33+1}\text{.35}}{3}\\ =1.326\\ =1.33\end{array}$

Conclusion:

Thus, the ratio of neutrons and protons for the stable isotope of silver is $1.30$.

The ratio of neutrons and protons for the stable isotope of cadmium is $1.33$.

The ratio of neutrons and protons for the stable isotope of indium is $1.35$.

The average value of ratio is $1.33$.

(e)

To determine

The neutron and proton numbers for Thorium (Th), Palladium (Pa), and Uranium (U) and the ratio of neutrons and protons for the stable isotopes of Thorium, Palladium and Uranium.

Answer to Problem 1SP

The neutron number for Thorium (Th) is 142 and proton number of Thorium is 90.

The neutron number for Palladium (Pa) is 140 and proton number of Palladium is 91.

The neutron number for Uranium (U) is 146 and proton number of Uranium is 92.

Thus, the ratio of neutrons and protons for the stable isotope of Thorium is $1.58$.

The ratio of neutrons and protons for the stable isotope of Palladium is $1.54$.

The ratio of neutrons and protons for the stable isotope of Uranium is $1.59$.

The average value of ratio is $1.57$.

Explanation of Solution

From the periodic table, Atomic number of Thorium is $90$ and mass number is $232$.

Therefore number proton number is $90$.

Therefore neutron number is $232-90=142$.

From the periodic table, Atomic number of Palladium is $91$ and mass number is $231$.

Therefore number proton number is $91$ .

Therefore neutron number is $231-91=140$.

From the periodic table, Atomic number of Uranium is $92$ and mass number is $238$.

Therefore number proton number is $92$.

Therefore neutron number is $238-92=146$.

The stable isotopes of Thorium, Palladium, and Uranium are Thorium-232, Palladium-231, and Uranium-$238$ respectively.

In the case of $\text{Thorium-232}$, the neutron number is 142 and proton number is $90$.

Write the expression for the ratio of neutron to proton of isotope.

$\text{Ratio}\text{=}\frac{N_n}{N_p}$ . (1)

Substitute $142$ for $N_n$ and $90$ for $N_p$ in the above equation to get $\text{ratio of neutron to proton}$.

$\begin{array}{c}\text{Ratio}\text{=}\frac{142}{90}\\ =1.58\end{array}$

In the case of $\text{Palladium-231}$, the neutron number is $140$  and proton number is $91$.

Substitute $140$  for $N_n$ and $91$  for $N_p$ in equation (1) to get ratio of neutron to proton for $\text{Palladium-231}$.

$\begin{array}{c}\text{Ratio}=\frac{140}{91}\\ =1.53\end{array}$

In the case of $\text{Uranium-238}$, the neutron number is $146$  and proton number is $92$.

Substitute $146$ for $N_n$ and $92$ for $N_p$ in the above equation to get ratio of neutron to proton for $\text{Uranium-238}$.

$\begin{array}{c}\text{Ratio}=\frac{146}{92}\\ =1.59\end{array}$

Write the average value of ratios.

$\begin{array}{c}\text{average}\text{=}\frac{\text{1}\text{.58+1}\text{.53+1}\text{.59}}{3}\\ =1.57\end{array}$

Conclusion:

Thus, the neutron number for Thorium (Th) is 142 and proton number of Thorium is 90.

The neutron number for Palladium (Pa) is 140 and proton number of Palladium is 91.

The neutron number for Uranium (U) is 146 and proton number of Uranium is 92.

The ratio of neutrons and protons for the stable isotope of Thorium is $1.58$.

The ratio of neutrons and protons for the stable isotope of Palladium is $1.54$.

The ratio of neutrons and protons for the stable isotope of Uranium is $1.59$.

The average value of ratio is $1.57$.

(f)

To determine

Why there are extra neutrons when uranium or thorium undergo fission by comparing the ratios of parts b, d, and e.

Answer to Problem 1SP

The ratio of neutron to proton for heavy nuclei are very large compared to medium nuclei and light nuclei. Therefore during fission these heavy nuclei have to produce extra neutrons to get stable medium nuclei or lighter nuclei.

Explanation of Solution

The neutron to proton ratio simply gives the idea about the extra number of neutron present in the nucleus.

The neutron to proton ratio for heavy nuclei is around $1.57$ and that of medium nuclei is $1.33$ and that of lighter nuclei is $1.0$. This data indicates that stable nucleus of a lighter one requires equal amount of neutron and proton, but stable nucleus of medium nucleus requires more neutron than protons, but stable nucleus of heavier nucleus requires very large amount of neutrons.

Therefore during fission of heavier to medium nuclei requires emission of extra neutron to get stable nuclei.

Conclusion:

Thus, the ratio of neutron to proton for heavy nuclei are very large compared to medium nuclei and light nuclei. Therefore during fission these heavy nuclei have to produce extra neutrons to get stable medium nuclei or lighter nuclei.