Part of one row of the periodic table is shown.
4 blue boxes in a row. The first has S n at the center and 50 above with tin and 118.71 below. The second has S b at the center and 51 above with antimony and 121.76 below. The third has T e at the center and 52 above with tellurium and 127.60 below. The fourth has I at the center and 53 above with iodine and 126.90 below.
Which element’s atoms have the greatest average number of neutrons?
tin (Sn)
antimony (Sb)
tellurium (Te)
iodine (I)
The number of neutrons in an element's atom is calculated by subtracting the atomic number (number of protons) from the atomic mass (average mass of the element's isotopes).
For tin (Sn), the atomic number is 50 and the atomic mass is 118.71, so the number of neutrons is 118.71 - 50 = 68.71.
For antimony (Sb), the atomic number is 51 and the atomic mass is 121.76, so the number of neutrons is 121.76 - 51 = 70.76.
For tellurium (Te), the atomic number is 52 and the atomic mass is 127.60, so the number of neutrons is 127.60 - 52 = 75.60.
For iodine (I), the atomic number is 53 and the atomic mass is 126.90, so the number of neutrons is 126.90 - 53 = 73.90.
Out of these four elements, tellurium (Te) has the greatest average number of neutrons.