The radioactive isotope `._(27)^(60)Co` which has now replaced radium in the treatment of cancer can be made by `a(n,p)` or `(n, gamma)` reaction. For each reaction, indicate the appropriate target nucleus. If the half life of `._(27)^(60)Co` is 7 year evaluate the decay constant in `s^(-1)`.

Explanation

To solve the problem, we will break it down into two parts: identifying the target nuclei for the reactions and calculating the decay constant for the radioactive isotope \( _{27}^{60}\text{Co} \).

### Part 1: Identifying the Target Nuclei

1. **For the (n, p) reaction:**
- The reaction involves a neutron being absorbed by a target nucleus, which results in the emission of a proton and the formation of the cobalt isotope.
- The appropriate target nucleus for this reaction is Nickel (Ni), specifically \( _{28}^{60}\text{Ni} \).
- The reaction can be written as:
\[ _{28}^{60}\text{Ni} + n \rightarrow _{27}^{60}\text{Co} + p \]

2. **For the (n, γ) reaction:**
- In this reaction, a neutron is absorbed by a target nucleus, and a gamma photon is emitted.
- The appropriate target nucleus for this reaction is also Nickel, specifically \( _{27}^{59}\text{Co} \).
- The reaction can be written as:
\[ _{27}^{59}\text{Co} + n \rightarrow _{27}^{60}\text{Co} + \gamma \]

### Part 2: Calculating the Decay Constant

The decay constant (\( \lambda \)) can be calculated using the half-life (\( t_{1/2} \)) of the isotope. The formula to relate half-life and decay constant is:
\[ \lambda = \frac{0.693}{t_{1/2}} \]

Given that the half-life of \( _{27}^{60}\text{Co} \) is 7 years, we first need to convert this time into seconds.

1. **Convert years to seconds:**
- 1 year = 365 days
- 1 day = 24 hours
- 1 hour = 60 minutes
- 1 minute = 60 seconds

Therefore, the total number of seconds in 7 years is:
\[ t_{1/2} = 7 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]
\[ t_{1/2} = 7 \times 365 \times 24 \times 60 \times 60 = 220752000 \text{ seconds} \]

2. **Calculate the decay constant:**
\[ \lambda = \frac{0.693}{220752000} \approx 3.14 \times 10^{-9} \text{ s}^{-1} \]

### Final Answers
- The target nucleus for the (n, p) reaction is \( _{28}^{60}\text{Ni} \).
- The target nucleus for the (n, γ) reaction is \( _{27}^{59}\text{Co} \).
- The decay constant \( \lambda \) is approximately \( 3.14 \times 10^{-9} \text{ s}^{-1} \).