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Question

An imaginary radioactive element decays to form elements X1,X2,X3,X4,X5 and X6, by ejecting two β-particles, followed by an α-particle, and again two (β-particles followed by an α -particle. What are the mass number and the atomic number of X6?

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Solution

The symbolic representation of the reaction is:
9292Xβparticle−−−−−X1βparticle−−−−−X2αparticle−−−−−X3βparticle−−−−−X4βparticle−−−−−X5αparticle−−−−−X6
In complete reaction, the total number of emitted a-particles = 2.
Loss of the total mass number due to two α-particles = 4 + 4 = 8
Loss of atomic number due to two α -particles = 2 + 2 = 4
In a complete reaction, the total number of emitted β - particles = 4.
Loss of mass number due to four β -particles = 0
Gain of atomic number due to four βparticles=1×4=4
Total loss of mass number = 8
Total loss or gain of atomic number
= 4 - 4 = 0
Mass number of X6=(2358)=227
Atomic number of X6=(920)=92


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