Question

27
iodine - 131 undergoes radioactive decay to form an isotope with 51 protons and 76 neutrons.
what type of decay is this? (1 point)
beta decay
lambda decay
gamma decay
alpha decay

Explanation:

• Beta decay: In beta decay, a neutron is converted into a proton and an electron (beta particle). This increases the atomic number (number of protons) by 1.
• Iodine - 131 has 53 protons (atomic number of iodine is 53). The daughter isotope has 51 protons. Wait, no! Wait, let's re - check. Wait, no, there is a mistake. Wait, iodine has atomic number 53. If after decay, the new element has 51 protons (atomic number 51 is for antimony, but wait, no, wait, no. Wait, no, in beta decay, \(n

ightarrow p + e^{-}\). The mass number of iodine - 131 is 131 (protons + neutrons = 53 + 78 = 131). The daughter isotope has protons \(p = 51\), neutrons \(n=76\), mass number \(A=p + n=51 + 76 = 127\). Wait, no, in beta decay, the general equation is \(^{A}_{Z}X
ightarrow^{A}_{Z + 1}Y+\beta^{-}+\bar{
u}\). Wait, no, another approach:

• Alpha decay: \(^{A}_{Z}X

ightarrow^{A - 4}_{Z-2}Y+\alpha\). If it were alpha decay from \(^{131}_{53}I\), the daughter would have \(Z = 53-2=51\) (which matches the problem, as the new element has 51 protons) and \(A=131 - 4=127\) (protons + neutrons \(51+76 = 127\)).

Answer:

Alpha decay