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Polonium ({ }_{84}^{210} {Po}) emits { }_{2}^{4} {He} particle

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Solving time: 4 mins

Polonium $(_{84}^{210} Po)$ emits $_{2}^{4} He$ particles and is converted into lead $(_{82}^{206} Pb)$ . This reaction is used for producing electric power in a space mission. $Po^{210}$ has half-life of $138.6$ days. Assuming an efficiency of $10%$ for the thermoelectric machine, how much $^{210} Po$ is required to produce $1.2 \times 1 0^{7} J$ of electric energy per day at the end of 693 days. Also find the initial activity of the material. Given : Masses of nuclei $^{210} Po = 209.98264 amu,^{206} Pb = 205.97440 amu,_{2}^{4} He = 4.00260 amu,$ $1 amu = 931 MeV / c^{2}$ and Avogadro's number $= 6 \times 1 0^{23} / mol$

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Optics And Modern Physics (DC Pandey)

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Mila Discussed

Polonium

(_{84}^{210} Po)

emits

_{2}^{4} He

particles and is converted into lead

(_{82}^{206} Pb)

. This reaction is used for producing electric power in a space mission.

Po^{210}

has half-life of

138.6

days. Assuming an efficiency of

10%

for the thermoelectric machine, how much

^{210} Po

is required to produce

1.2 \times 1 0^{7} J

of electric energy per day at the end of 693 days. Also find the initial activity of the material. Given : Masses of nuclei

^{210} Po = 209.98264 amu,^{206} Pb = 205.97440 amu,_{2}^{4} He = 4.00260 amu,

1 amu = 931 MeV / c^{2}

and Avogadro's number

= 6 \times 1 0^{23} / mol

13 mins ago

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Text solutionVerified

_{84}^{210} Po ⟶_{82}^{206} Pb +_{2}^{4} He Δ m = 0.00564 amu

Energy liberated per reaction

= (Δ m) 931 MeV

= 8.4 \times 1 0^{- 13} J

Electrical energy produced

= 8.4 \times 1 0^{- 14} J

Let

m g

^{210} Po

is required to produce the desired energy.

N λ (- \frac{d N}{d t}) = \frac{m}{210} \times 6 \times 1 0^{23} = \frac{0.693}{t _{1/2}} = 0.005 per day = λ N = \frac{( 0.005 ) ( 6 \times 1 0 ^{23} m )}{210} per day

t h ere f ore,

Electrical energy produced per day

= \frac{( 0.005 ) ( 6 \times 1 0 ^{23} m )}{210} \times 8.4 \times 1 0^{- 14} J

This is equal to

1.2 \times 1 0^{7} J

(given)

∴ m = 10 g

Activity at the end of 693 days is

R = \frac{0.005 \times 6 \times 1 0 ^{23} \times 10}{210} = \frac{1 0 ^{21}}{7} per day = R_{0} (\frac{1}{2})^{n}

Here,

n =

number of half-lives

= \frac{693}{138.6} = 5

∴ R_{0} = R (2)^{5} = 32 \times \frac{1 0 ^{21}}{7} = 4.57 \times 1 0^{21} per day

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Question Text	Polonium $(_{84}^{210} Po)$ emits $_{2}^{4} He$ particles and is converted into lead $(_{82}^{206} Pb)$ . This reaction is used for producing electric power in a space mission. $Po^{210}$ has half-life of $138.6$ days. Assuming an efficiency of $10%$ for the thermoelectric machine, how much $^{210} Po$ is required to produce $1.2 \times 1 0^{7} J$ of electric energy per day at the end of 693 days. Also find the initial activity of the material. Given : Masses of nuclei $^{210} Po = 209.98264 amu,^{206} Pb = 205.97440 amu,_{2}^{4} He = 4.00260 amu,$ $1 amu = 931 MeV / c^{2}$ and Avogadro's number $= 6 \times 1 0^{23} / mol$
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	114

Text solutionVerified

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