55

25Mn can be prepared by electron capture from:
(a) 55/24Cr
(b) 55/26Fe
(c) 56/25Mn
(d) 51/23V
(e) 56/26Fe

wouldn't it take 25 protons and one electron to make 24 protons?

so the answer is A?

To determine which option can be used to prepare 25Mn through electron capture, we need to understand the concept of electron capture and the change in atomic number and mass number during this process.

Electron capture is a nuclear reaction in which an electron from the inner electron shell is captured by a nucleus, resulting in the transformation of a proton into a neutron. As a result, the atomic number decreases by 1, while the mass number remains the same.

Let's analyze each option:

(a) 55/24Cr: Electron capture on chromium-55 (55/24Cr) would result in the formation of manganese-55 (55/25Mn). The atomic number does not change, so this option is not correct.

(b) 55/26Fe: Electron capture on iron-55 (55/26Fe) would lead to the formation of manganese-55 (55/25Mn). The atomic number does not change, so this option is not correct either.

(c) 56/25Mn: This option suggests that electron capture occurs on manganese-56 (56/25Mn) to form manganese-55 (55/25Mn). In this case, the atomic number decreases by 1, which is consistent with the process of electron capture. Therefore, this option is the correct one.

(d) 51/23V: Electron capture on vanadium-51 (51/23V) would result in the formation of titanium-51 (51/22Ti). The atomic number decreases, but the mass number also decreases, which contradicts the process of electron capture.

(e) 56/26Fe: This option suggests that electron capture takes place on iron-56 (56/26Fe) to form manganese-56 (56/25Mn). The change in atomic number does not match the process of electron capture, so this option is incorrect.

Therefore, the correct answer is option (c) 56/25Mn.