Q. 4.33 - An Introduction to General, Organic, and Biological Chemistry

Question

What are the number of protons, neutrons, and electrons in the following isotopes?

a. ${}_{38}{}^{89}Sr$

b. ${}_{24}{}^{52}Cr$

c. ${}_{16}{}^{34}S$

d. ${}_{35}{}^{81}Br$

Step-by-Step Solution

Answer

(a) No. of protons or electrons are $38$ and No. of neutrons are $51$ 

(b) No. of protons or electrons are $24$ and No. of neutrons are $28$

(c) No. of protons or electrons are $16$ and No. of neutrons are $18$

(d) No. of protons or electrons are $35$ and No. of neutrons are $46$

Part (a) Step 1: Given information

We need to find out the no. of protons, neutrons and electrons in ${}_{38}{}^{89}Sr$

Part (a) Step 2: Simplify

We know that the atom in standard form is represented as, ${}_Z{}^{A}X$

Where, $Z=$Atomic no. or no. of protons or no. of electrons

And $A=$Mass no.

And $X=$Symbol of the element

As we have, ${}_{38}{}^{89}Sr$

So, here

$$\begin{gathered} A=89 \\ Z=38 \end{gathered}$$

Therefore,

No. of protons or no. of electrons are, $38$

And no. of neutrons are, $A-Z=89-38=51$

Part (b) Step 1: Given information

We need to find out the no. of protons, neutrons and electrons in ${}_{24}{}^{52}Cr$

Part (b) Step 2: Simplify

We know that the atom in standard form is represented as, ${}_Z{}^{A}X$

Where, $Z=$Atomic no. or no. of protons or no. of electrons

And $A=$Mass no.

And $X=$Symbol of the element

As we have, ${}_{24}{}^{52}Cr$

So, here

$$\begin{gathered} A=52 \\ Z=24 \end{gathered}$$

Therefore,

No. of protons or no. of electrons are, $24$

And no. of neutrons are, $A-Z=52-24=28$

Part (c) Step 1: Given information

We need to find out the no. of protons, neutrons and electrons in ${}_{16}{}^{34}S$

Part (c) Step 2: Simplify

We know that the atom in standard form is represented as, ${}_Z{}^{A}X$

Where, $Z=$Atomic no. or no. of protons or no. of electrons

And $A=$Mass no.

And $X=$Symbol of the element

As we have, ${}_{16}{}^{34}S$

So, here

$$\begin{gathered} A=34 \\ Z=16 \end{gathered}$$

Therefore,

No. of protons or no. of electrons are, $16$

And no. of neutrons are, $A-Z=34-16=18$

Part (d) Step 1: Given information

We need to find out the no. of protons, neutrons and electrons in ${}_{35}{}^{81}Br$

Part (d) Step 2: Simplify

We know that the atom in standard form is represented as, ${}_Z{}^{A}X$

Where, $Z=$Atomic no. or no. of protons or no. of electrons

And $A=$Mass no.

And $X=$Symbol of the element

As we have, ${}_{35}{}^{81}Br$

So, here

$$\begin{gathered} A=81 \\ Z=35 \end{gathered}$$

Therefore,

No. of protons or no. of electrons are, $35$

And no. of neutrons are, $A-Z=81-35=46$