Consider the following nuclide symbol: 51V4+ How many neutrons does this nuclide have? How many Electrons?

Question

If someone could please explain how to find the neutrons and electrons from this nuclide symbol that would be really helpful. Thank you so much.

Answer 1

The Format for isotopes is _Z^AX where A is the atomic mass number = n_p + n_n , Z is the atomic number = n_p and the overall charge is equal to the n_p-n_e.

You are given V⁴⁺- 51 (You are not given Z, but it is unnecessary because V has a Z of 23.

Z=n_p=23 and n_n = 51- 23 = 28 and the n_e = n_p-4 = 19

You should consider a tutor. Take care.

Answer 2

Please check the formatting of your question. Assuming 51 is the upper left superscript and 4+ is the upper right superscript and V is the element symbol:

a) you must subtract the atomic number of vanadium (23, you can find on the periodic table), from the mass number in the upper left superscript. (51).

b) You must subtract the charge number in the upper right superscript from the atomic number (which is usually placed in the lower left subscript or left blank if the element's atomic symbol is given). 4+ is the excess amount of protons compared to electrons.