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Under Certain Circumstances, a Nucleus Can Decay by Emitting a Particle More Massive than an α-particle. Consider the Following Decay Processes: - CBSE (Science) Class 12 - Physics

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Question

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

`""_88^223Ra -> ""_82^209Pb + _6^14C`

`""_88^223Ra -> ""_86^219Rn + _2^4 He`

Calculate the Q-values for these decays and determine that both are energetically allowed.

Solution

Take a `""_6^14C` emission nuclear reaction:

`""_88^223Ra -> _82^209Pb + _6^14C`

We know that:

Mass of `""_88^223Ra` m1 = 223.01850 u

Mass of `""_82^209Pb` m2 = 208.98107 u

Mass of `""_6^14C`, m3 = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = (m1 − m2 − m3c2

= (223.01850 − 208.98107 − 14.00324) c2

= (0.03419 c2) u

But 1 u = 931.5 MeV/c2

Q = 0.03419 × 931.5

= 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take a `""_2^4He` emission nuclear reaction:

`""_88^223Ra -> _86^229Rn + _2^4He`

We know that:

Mass of `""_88^223Ra`, m1 = 223.01850

Mass of `""_82^219Rn`  m2 = 219.00948

Mass of `""_2^4He`, m3 = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m1 − m2 − m3c2

= (223.01850 − 219.00948 − 4.00260) C2

= (0.00642 c2) u

= 0.00642 × 931.5 = 5.98 MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

  Is there an error in this question or solution?

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Solution Under Certain Circumstances, a Nucleus Can Decay by Emitting a Particle More Massive than an α-particle. Consider the Following Decay Processes: Concept: Radioactivity - Law of Radioactive Decay.
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