Answer :
Sure! Let's solve the problem step-by-step to identify the values of \( a \), \( b \), and \( c \) in the given table.
We are examining three isotopes of hydrogen: hydrogen-1 (\( H-1 \)), deuterium (\( H-2 \)), and tritium (\( H-3 \)).
First, we know the general characteristics of hydrogen isotopes:
- All isotopes of hydrogen have the same number of protons.
- All isotopes of hydrogen have the same number of electrons.
- The number of neutrons differentiates the isotopes.
Given:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Isotope} & \text{Number of Protons} & \text{Number of Electrons} & \text{Number of Neutrons} \\ \hline \text{[tex]$H -1$[/tex]} & 1 & 1 & 0 \\
\hline \text{[tex]$H -2$[/tex]} & a & 1 & b \\
\hline \text{[tex]$H -3$[/tex]} & 1 & 1 & c \\
\hline
\end{array}
\][/tex]
Step-by-step solution:
1. For \(H -1\) (Protium):
- Number of protons = 1
- Number of electrons = 1
- Number of neutrons = 0
2. For \(H -2\) (Deuterium):
- Number of protons should be the same as \(H-1\), i.e., \( a = 1 \).
- Number of electrons is already given as 1.
- Deuterium has one neutron, thus \( b = 1 \).
3. For \(H -3\) (Tritium):
- Number of protons should be the same as \(H-1\), so it remains 1.
- Number of electrons is already given as 1.
- Tritium has two neutrons, so \( c = 2 \).
Now, filling the values into the table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Isotope} & \text{Number of Protons} & \text{Number of Electrons} & \text{Number of Neutrons} \\ \hline \text{[tex]$H -1$[/tex]} & 1 & 1 & 0 \\
\hline \text{[tex]$H -2$[/tex]} & 1 & 1 & 1 \\
\hline \text{[tex]$H -3$[/tex]} & 1 & 1 & 2 \\
\hline
\end{array}
\][/tex]
Therefore, the values of \( a \), \( b \), and \( c \) are:
[tex]\[ a = 1, \ b = 1, \ c = 2 \][/tex]
So, the correct option is:
D. [tex]\(a=1, b=1, c=2\)[/tex]
We are examining three isotopes of hydrogen: hydrogen-1 (\( H-1 \)), deuterium (\( H-2 \)), and tritium (\( H-3 \)).
First, we know the general characteristics of hydrogen isotopes:
- All isotopes of hydrogen have the same number of protons.
- All isotopes of hydrogen have the same number of electrons.
- The number of neutrons differentiates the isotopes.
Given:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Isotope} & \text{Number of Protons} & \text{Number of Electrons} & \text{Number of Neutrons} \\ \hline \text{[tex]$H -1$[/tex]} & 1 & 1 & 0 \\
\hline \text{[tex]$H -2$[/tex]} & a & 1 & b \\
\hline \text{[tex]$H -3$[/tex]} & 1 & 1 & c \\
\hline
\end{array}
\][/tex]
Step-by-step solution:
1. For \(H -1\) (Protium):
- Number of protons = 1
- Number of electrons = 1
- Number of neutrons = 0
2. For \(H -2\) (Deuterium):
- Number of protons should be the same as \(H-1\), i.e., \( a = 1 \).
- Number of electrons is already given as 1.
- Deuterium has one neutron, thus \( b = 1 \).
3. For \(H -3\) (Tritium):
- Number of protons should be the same as \(H-1\), so it remains 1.
- Number of electrons is already given as 1.
- Tritium has two neutrons, so \( c = 2 \).
Now, filling the values into the table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Isotope} & \text{Number of Protons} & \text{Number of Electrons} & \text{Number of Neutrons} \\ \hline \text{[tex]$H -1$[/tex]} & 1 & 1 & 0 \\
\hline \text{[tex]$H -2$[/tex]} & 1 & 1 & 1 \\
\hline \text{[tex]$H -3$[/tex]} & 1 & 1 & 2 \\
\hline
\end{array}
\][/tex]
Therefore, the values of \( a \), \( b \), and \( c \) are:
[tex]\[ a = 1, \ b = 1, \ c = 2 \][/tex]
So, the correct option is:
D. [tex]\(a=1, b=1, c=2\)[/tex]
