(A) Cobalt-$60$ $(^{60}Co)$ is a radioactive isotope of cobalt.It undergoes radioactive decay and emits high-energy $\gamma$-radiations.These $\gamma$

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(A) Cobalt-$60$ $(^{60}Co)$ is a radioactive isotope of cobalt.It undergoes radioactive decay and emits high-energy $\gamma$-radiations.These $\gamma$-radiations have high penetrating power and are used in radiation therapy to target and destroy cancerous cells.Therefore,the Assertion is correct,the Reason is correct,and the Reason provides the correct explanation for the Assertion.

Class 12 Physics Nuclei Properties of Alpha, Beta and Gamma Rays and Decay Process

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Assertion: Cobalt-$60$ is useful in cancer therapy.
Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

AIIMS 2006 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

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Nuclei

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Properties of Alpha, Beta and Gamma Rays and Decay Process

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AIIMS 2006 Paper All AIIMS years →

$A$ nucleus $_Z{X^A}$ emits $3 \alpha$-particles and $5 \beta$-particles. The ratio of total neutrons to protons in the final nucleus is:

Medium

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In the given nuclear reaction,$A, B, C, D, E$ represent:
$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

Medium

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In medicine,to destroy cancer cells . . . . . . rays are used.

EasyGUJCET 2017

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$A$ nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

MediumAIPMT 2011

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Identify the particle $x$ in the following reaction: ${}_3^7Li(p, x){}_4Be^8$,where $p$ is a proton.

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Assertion: Cobalt-$60$ is useful in cancer therapy.Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

Similar Questions

$A$ nucleus $_Z{X^A}$ emits $3 \alpha$-particles and $5 \beta$-particles. The ratio of total neutrons to protons in the final nucleus is:

In the given nuclear reaction,$A, B, C, D, E$ represent:$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

In medicine,to destroy cancer cells . . . . . . rays are used.

$A$ nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

Identify the particle $x$ in the following reaction: ${}_3^7Li(p, x){}_4Be^8$,where $p$ is a proton.

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Assertion: Cobalt-$60$ is useful in cancer therapy.
Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

In the given nuclear reaction,$A, B, C, D, E$ represent:
$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$