WBJEE · Chemistry · Chemical Kinetics
The nucleus \(^{64}_{29}Cu\) accepts an orbital clectron to yield.
- A \(\frac{65}{28} \mathrm{Ni}\)
- B \(\frac{64}{30} \mathrm{Zn}\)
- C \(\frac{64}{28} \mathrm{Ni}\)
- D \(\frac{65}{30} \mathrm{zn}\)
Answer & Solution
Correct Answer
(C) \(\frac{64}{28} \mathrm{Ni}\)
Step-by-step Solution
Detailed explanation
The nucleus \(^{64}_{29}Cu\) accepts an orbital electron to yield \(^{64}_{28}Ni\). The atomic number of \(\mathrm{Cu}\) is 29 , which is equal to the number of electrons and also equal to the number of protons. When \(^{64}_{29}Cu\) accepts an orbital electron then electrons…
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