The kinetic energy in keV of the alpha particle- when the nucleus 21084 Po at rest undergoes alpha decay- is

^210_84 Po = ^206_82 Pb + ^4_2 He + Δ E m ( ^206_82 Pb ) = 205.974455 u m ( ^4_2 He ) = 4.002603 u m ( ^206_82 Pb ) + m ( ^4_2 He ) = 209.977058 u = 0.00 ...

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Standard XII

Physics

Half Life and Average Life

The kinetic e...

Question

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} P o$ at rest undergoes alpha decay, is

5707

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5818

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5422

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5319

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Solution

The correct option is D $5319$
$_{84}^{210} P o =_{82}^{206} P b +_{2}^{4} H e + Δ E$
$m (_{82}^{206} P b) = 205.974455 u$
$m (_{2}^{4} H e) = 4.002603 u$
$m (_{82}^{206} P b) + m (_{2}^{4} H e) = 209.977058 u = 0.005818 u$
$Δ E = 0.005818 \times 931.5$
$= 5.419467 meV = 5419.467 MeV = 5419.5 MeV$
$Δ E > K E_{α}$

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The kinetic energy (in keV) of the alpha particle, when the nucleus 21084Po at rest undergoes alpha decay, is

The correct option is D 531921084Po=20682Pb+42He+ΔE m(20682Pb)=205.974455 u m(42He)=4.002603 u m(20682Pb)+m(42He)=209.977058 u=0.005818 u ΔE=0.005818×931.5 =5.419467 meV=5419.467 MeV=5419.5 MeV ΔE>KEα

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} P o$ at rest undergoes alpha decay, is