Q17 / 155 · WBJEE 2018 · Chemistry

The nucleus $^{64}_{29}Cu$ accepts an orbital clectron to yield.

  • A
    $\frac{65}{28} \mathrm{Ni}$
  • B
    $\frac{64}{30} \mathrm{Zn}$
  • Correct answer
    C
    $\frac{64}{28} \mathrm{Ni}$
  • D
    $\frac{65}{30} \mathrm{zn}$

Solution

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 subtract from the atomic number, i.e. $29-1=28$ $$ ^{64}_{29}Cu+{ }_{-1} e^{0} \longrightarrow{ }_{28}^{64} \mathrm{Ni} $$

Hence, the correct answer is option: 'C'.

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