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Vika [28.1K]
3 years ago
6

A large atom decays and emits a particle. After the reaction is complete, the atom’s mass has changed substantially. What kind o

f particle has most likely been emitted?
alpha particle
beta particle
positron
photon
Physics
2 answers:
Montano1993 [528]3 years ago
5 0

Answer:

alpha particle

Explanation:

In a process where alpha particle is emitted, a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}+_2^4\alpha

The atomic mass of the resulting lighter nuclei reduces by 4.

The process in which beta particle is emitted, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The mass of the resultant nuclei remains same.

The process in which a proton gets converted to neutron and an electron neutrino, a positron is released. This particle carries a charge of +1 units.

_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

The mass of the resultant nuclei remains unchanged.

In gamma ray emission,

_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma

A photon is emitted. An unstable nuclei gets stabilized by release of photon. There is no change in the nuclei.

Thus, in a reaction of decay of a large atom, release of alpha particle would result in substantial change in atomic's mass.

LiRa [457]3 years ago
3 0
The answer your looking for is, A. alpha particle.
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Answer:

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   From the question we are told that

           The height is h_s = 30 \ cm

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So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

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              v^2 = \frac{10gh_s}{7}

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   The moment of inertial is different for circle and it is mathematically represented as

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Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

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Answer:

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