# Under Certain Circumstances, a Nucleus Can Decay by Emitting a Particle More Massive than an α-particle. Consider the Following Decay Processes: - Physics

Numerical

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.

#### Solution

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

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

We know that:

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

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

Mass of ""_6^14"C", m3 = 14.00324 u

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

Q = (m1 − m2 − m3) c2

= (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 ""_2^4"He" emission nuclear reaction:

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

We know that:

Mass of ""_88^223"Ra", m1 = 223.01850

Mass of ""_82^219"Rn"  m2 = 219.00948

Mass of ""_2^4"He", m3 = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m1 − m2 − m3) c2

= (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.

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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.26 | Page 464
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 26 | Page 464

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