Answer
a). The correct answer is (1) greater, because $k_{min}=(1+\frac{m_{a}}{M_{A}})|Q|$. The difference in Q value is greater than the difference caused by the target mass (15 and 20).
b). 3.05
Work Step by Step
a). The correct answer is (1) greater, because $k_{min}=(1+\frac{m_{a}}{M_{A}})|Q|$. The difference in Q value is greater than the difference caused by the target mass (15 and 20).
b). $k_{min1}=(1+\frac{m_{a1}}{M_{A1}})|Q1|$
$k_{min2}=(1+\frac{m_{a2}}{M_{A2}})|Q2|$
So, $\frac{k_{min1}}{k_{min2}}=\frac{(1+\frac{m_{a1}}{M_{A1}})|Q1|}{(1+\frac{m_{a2}}{M_{A2}})|Q2|}=\frac{(1+\frac{1}{15})}{(1+\frac{1}{20})}\times3=3.05$
