The kinetic energy of the - - particle emitted in the decay 23894Pu - 23492U - 42He-The atomic masses of 238Pu- 234U and - particle are 238-04955u- 234-04095 u and 4-002603u respectively- Neglect any recoil of the nucleus

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The kinetic e...

Question

The kinetic energy of the $α$ - particle emitted in the decay $_{94}^{238} P u \to_{92}^{234} U +$ $_{2}^{4} H e$ .
(The atomic masses of $^{238} P u,^{234} U$ and $α$ particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)

4.392 MeV

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6.259 MeV

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5.592 MeV

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4.952 MeV

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Solution

The correct option is C 5.592 MeV
Assuming that energy librated by mass defect is converted into KE of $α -$ particle.
$Δ m = (238.04955 - 234.04095 - 4.002603) μ$
$= 0.005997 μ$
$K E = E = 0.005997 \times 931.478 M e V$
$K E = 5.586 M e V$

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The kinetic energy of the α - particle emitted in the decay 23894Pu→23492U+ 42He.(The atomic masses of 238Pu,234U and α particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)

The correct option is C 5.592 MeVAssuming that energy librated by mass defect is converted into KE of α−particle.Δm=(238.04955−234.04095−4.002603)μ =0.005997μKE=E=0.005997×931.478 MeVKE=5.586 MeV

The kinetic energy of the $α$ - particle emitted in the decay $_{94}^{238} P u \to_{92}^{234} U +$ $_{2}^{4} H e$ .
(The atomic masses of $^{238} P u,^{234} U$ and $α$ particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)

The correct option is C 5.592 MeV
Assuming that energy librated by mass defect is converted into KE of $α -$ particle.
$Δ m = (238.04955 - 234.04095 - 4.002603) μ$
$= 0.005997 μ$
$K E = E = 0.005997 \times 931.478 M e V$
$K E = 5.586 M e V$