JEE Main 2015 · 11 Apr · Shift 1 · Physics Q28

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PhysicsSingle correct+4 / -1Question 28 of 90

Let $N_\beta$ be the number of β particles emitted by 1 gram of $^{24}$ Na radioactive nuclei (half life = 15 hrs) in 7.5 hours, $N_\beta$ is close to (Avogadro number = $6.023\times10^{23}$ /g. mole) :

6.2\times10^{21}

7.5\times10^{21}

1.25\times10^{22}

1.75\times10^{22}

Answer and worked solution

Official answer: B

$N_0=\dfrac{6.023\times10^{23}}{24}=2.51\times10^{22}$ . In half a half-life the fraction decayed is $1-2^{-1/2}=0.293$ :
$N_\beta=0.293\times2.51\times10^{22}\approx7.5\times10^{21}.$

Topics:Radioactive decay

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 28