Plutonium has atomic mass 210 and a decay constant equal to 5-8- 10-8s-1- The number of -particles emitted per second by 1 mg plutonium is Avogadro-s constant -6-0- 1023

The correct option is A 1.7× 10^11 No. of atoms in 1mg of Plutonium =10^ 3210× N_A N = 2.85 × 10^18 First order decay: dNdt = λ N dNdt ...

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Standard XII

Chemistry

Heat of Reaction

Plutonium has...

Question

Plutonium has atomic mass $210$ and a decay constant equal to $5.8 \times 10^{- 8} s^{- 1}$ . The number of $α - p a r t i c l e s$ emitted per second by $1 m g$ plutonium is (Avogadro's constant $= 6.0 \times 10^{23})$

1.7 \times 10^{9}

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1.7 \times 10^{11}

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2.9 \times 10^{11}

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3.4 \times 10^{9}

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Solution

The correct option is A $1.7 \times 10^{11}$
No. of atoms in 1mg of Plutonium $= \frac{10^{- 3}}{210} \times N_{A}$
$N = 2.85 \times 10^{18}$
First order decay:
$\frac{d N}{d t} = - λ N$
$\frac{d N}{d t} = - 5.8 \times 10^{- 8} \times (2.85 \times 10^{18})$
$\frac{d N}{d t} = 1.66 \times 10^{11} s^{- 1}$

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Plutonium has atomic mass 210 and a decay constant equal to 5.8×10−8s−1. The number of α−particles emitted per second by 1mg plutonium is (Avogadro's constant =6.0×1023)

The correct option is A 1.7×1011No. of atoms in 1mg of Plutonium =10−3210×NAN=2.85×1018First order decay:dNdt=−λNdNdt=−5.8×10−8×(2.85×1018)dNdt=1.66×1011s−1

Plutonium has atomic mass $210$ and a decay constant equal to $5.8 \times 10^{- 8} s^{- 1}$ . The number of $α - p a r t i c l e s$ emitted per second by $1 m g$ plutonium is (Avogadro's constant $= 6.0 \times 10^{23})$