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

The correct option is A 5319 ^210_84Po→_2^4He+_82^206 Pb Q=(209.982876 4.002603 205.97455)C^2 #160; #160; =5.422MeV From conservation ...

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Standard X

Physics

Nuclear Fission

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

5319

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Solution

The correct option is A $5319$
$_{84}^{210} P o \to_{2}^{4} H e +_{82}^{206}$ Pb
$Q = (209.982876 - 4.002603 - 205.97455) C^{2}$
$= 5.422 M e V$

From conservation of momentum:
$\sqrt{2 K_{1} (4)} = \sqrt{2 K_{2} (206)}$
$4 K_{1} = 206 K_{2}$
$K_{1} = \frac{103}{2} K_{2}$
$K_{1} + K_{2} = 5.422$
$K_{1} + \frac{2}{103} K_{1} = 5.422$
$\Rightarrow \frac{105}{103} K_{1} = 5.422$
$K_{1} = 5.319 M e V = 5319 K e V$

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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 A 531921084Po→42He+20682 PbQ=(209.982876−4.002603−205.97455)C2 =5.422MeVFrom conservation of momentum:√2K1(4)=√2K2(206)4K1=206K2K1=1032K2K1+K2=5.422K1+2103K1=5.422⇒105103K1=5.422K1=5.319MeV=5319KeV

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