Question

part of one row of the periodic table is shown. which elements atoms have the greatest average number of neutrons?
○ tin (sn)
○ antimony (sb)
○ tellurium (te)
○ iodine (i)
(image shows: sn (50, 118.71), sb (51, 121.76), te (52, 127.60), i (53, 126.90))

Explanation:

Step1: Recall neutron calculation formula

The number of neutrons in an atom is approximately calculated by subtracting the atomic number (Z) from the atomic mass (A), i.e., $n \approx A - Z$.

Step2: Calculate neutrons for Sn

For tin (Sn), atomic number $Z = 50$, atomic mass $A \approx 118.71$. So neutrons $n_{Sn} \approx 118.71 - 50 = 68.71$.

Step3: Calculate neutrons for Sb

For antimony (Sb), $Z = 51$, $A \approx 121.76$. Neutrons $n_{Sb} \approx 121.76 - 51 = 70.76$.

Step4: Calculate neutrons for Te

For tellurium (Te), $Z = 52$, $A \approx 127.60$. Neutrons $n_{Te} \approx 127.60 - 52 = 75.60$.

Step5: Calculate neutrons for I

For iodine (I), $Z = 53$, $A \approx 126.90$. Neutrons $n_{I} \approx 126.90 - 53 = 73.90$.

Step6: Compare neutron numbers

Compare $68.71$, $70.76$, $75.60$, $73.90$. We see that $75.60$ (Te) is the largest.

Answer:

tellurium (Te)