Difficulty: Easy
Correct Answer: γ-rays
Explanation:
Introduction:
Cobalt-60 (Co-60) is a man-made radioisotope produced by neutron activation of cobalt-59. Its decay scheme makes it extremely useful as a high-energy, high-intensity radiation source for teletherapy units and for non-destructive testing (NDT) in industry. Identifying its principal emission is fundamental to shielding design and safety protocols.
Given Data / Assumptions:
Concept / Approach:
While Co-60 undergoes beta decay, the practically important radiation for therapy and radiography is the cascade of two high-energy gamma photons (~1.17 MeV and ~1.33 MeV). These γ-rays are highly penetrating, enabling treatment of tumors at depth and radiographic inspection of thick components. X-rays are produced by electron transitions or Bremsstrahlung in X-ray tubes, not by Co-60 decay directly.
Step-by-Step Solution:
Recognize Co-60 decay: β- emission to excited Ni-60.Identify emitted photons: two γ-rays near 1.17 and 1.33 MeV.Select γ-rays as the dominant useful radiation from sealed Co-60 sources.
Verification / Alternative check:
Databases and shielding charts list Co-60 as a gamma emitter with average energies around 1.25 MeV, which dictate lead or concrete thicknesses for safe operation.
Why Other Options Are Wrong:
Common Pitfalls:
Equating the decay mode (beta) with the useful external radiation (gamma). In sealed sources, betas are shielded by the capsule; gammas escape and are used.
Final Answer:
γ-rays
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