Question

Cobalt-60 is used to irradiate tumors with $\gamma$-rays.
${}_{27}^{60}Co{⟶}_{28}^{60}Ni{+}_{-1}^{0}e+\gamma$
Which is responsible for destroying cancer-cells?

• A. $\gamma -rays$
• B. $\alpha -rays$
• C. $\beta -rays$
• D. None of these

Answer 1

The correct option is A $\gamma -rays$

${}_{27}^{60}Co{⟶}_{28}^{60}Ni{+}_{-1}^{0}e+\gamma$

An external radiation procedure is known as the gamma knife radiosurgery and involves focusing gamma radiation from $201$ sources of cobalt-$60$ sources on a precise area of the brain with a cancerous tumour. Worldwide, over $30,000$ patients are treated annually, generally as outpatients.