Question

researchers are using nickel-63 to develop small batteries that can generate electricity for up to fifty years. as nickel-63 decays, it only emits radiation that is attracted toward a positively charged metal plate. select the nuclear reaction that describes this process. \\(\ce{_{28}^{63}ni -> _{2}^{4}\alpha + _{-1}^{0}\beta + _{27}^{59}co}\\) \\(\ce{_{28}^{63}ni -> _{26}^{59}fe + _{2}^{4}he + _{0}^{0}\gamma}\\) \\(\ce{_{28}^{63}ni -> _{-1}^{0}e + _{29}^{63}cu}\\) \\(\ce{_{28}^{63}ni -> _{2}^{4}he + _{26}^{59}fe}\\)

Explanation:

Step1: Analyze Radiation Charge

The radiation is attracted to a positively charged plate, so it must be negatively charged. A beta particle ($\ce{^0_{-1}e}$ or $\ce{^0_{-1}\beta}$) has a negative charge.

Step2: Check Mass and Atomic Numbers

For a nuclear reaction, the sum of mass numbers (top) and atomic numbers (bottom) on both sides must be equal.

• For the reaction $\ce{^{63}_{28}Ni -> ^0_{-1}e + ^{63}_{29}Cu}$:
• Mass numbers: $63 = 0 + 63$ (balanced).
• Atomic numbers: $28 = -1 + 29$ (balanced).
• Other reactions:
• First option has multiple emissions, mass/atomic numbers don't balance.
• Second option has $\ce{^4_2He}$ (alpha, positive) and $\gamma$ (neutral), not negative radiation.
• Fourth option has alpha particle (positive), not negative.

Answer:

$\boldsymbol{\ce{^{63}_{28}Ni -> ^0_{-1}e + ^{63}_{29}Cu}}$ (the third option: $\ce{^{63}_{28}Ni -> ^0_{-1}e + ^{63}_{29}Cu}$)