Question

A free neutrons is unstable against $\beta$ - decay with a half life of about $600$ seconds-

• A. The expression of this decay process in $n\to p+c+\overline{v}$
• B. If there are $6\times {10}^{10}$ free neutrons initially, the time by Which $4.5\times {10}^{10}$ of them have decayed is $1200$ sec.
• C. They decay rate of the sample is $0.593$ Bq.
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A The expression of this decay process in $n\to p+c+\overline{v}$
B If there are $6\times {10}^{10}$ free neutrons initially, the time by Which $4.5\times {10}^{10}$ of them have decayed is $1200$ sec.

Beta particle is the high speed and high energy electron or positron particles. These electron or positrons are emitted during radioactive decay.

$n\to p+e+\overline{v}$

so option A is correct.

If $A_0=6\ast {10}^{10}$

After one half life (600 sec) $A_{t1}=3\ast {10}^{10}$

After second half life (2* 600 sec) $A_{t1}=1.5\ast {10}^{10}$

Total decay after second half-life (after 1200 sec)$=6\ast {10}^{10}-1.5\ast {10}^{10}=4.5\ast {10}^{10}$

Option B is correct.

Hence option 'A' & 'B' are correct.