A single electron in an ion head ionisation energy equal to 217.6 eV. What is the total number of neutrons present in one ion of it?
Answers
Answered by
12
Ionization energy = 13.6 x z^2
217.6 = 13.6 x z^2
z^2 = 217.6/13.6
z = 4
element = Be
neutrons = 4
217.6 = 13.6 x z^2
z^2 = 217.6/13.6
z = 4
element = Be
neutrons = 4
Answered by
15
Answer:
answer is 5
Explanation:
no. of neutrons = mass - protons
using energy formula we get Z=4
Z = no. of protons= 4---- Be
mass number of be = 9
no. of neutrons = 9-4
therefore no. of neutrons = 5
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