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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:

\[\ce{^223_88Ra -> ^209_82Pb + ^14_6C}\]

\[\ce{^223_88 Ra -> ^219_86 Rn + ^4_2He}\]

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

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#### Solution

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

\[\ce{^223_88Ra -> ^209_82Pb + ^14_6C}\]

We know that:

Mass of `""_88^223"Ra"` m_{1} = 223.01850 u

Mass of `""_82^209"Pb"` m_{2} = 208.98107 u

Mass of `""_6^14"C"`, m_{3} = 14.00324 u

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

Q = (m_{1} − m_{2} − m_{3}) c^{2}

= (223.01850 − 208.98107 − 14.00324) c^{2}

= (0.03419 c^{2}) u

But 1 u = 931.5 MeV/c^{2}

∴ 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 `""_2^4"He"` emission nuclear reaction:

\[\ce{^223_88 Ra -> ^219_86 Rn + ^4_2He}\]

We know that:

Mass of `""_88^223"Ra"`, m_{1} = 223.01850

Mass of `""_82^219"Rn"` m_{2} = 219.00948

Mass of `""_2^4"He"`, m_{3} = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m_{1} − m_{2} − m_{3}) c^{2}

= (223.01850 − 219.00948 − 4.00260) C^{2}

= (0.00642 c^{2}) 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.

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