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A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years.
The time, in years, after which one-fourth of the material remains is
The time, in years, after which one-fourth of the material remains is
- a)1080
- b)2430
- c)3240
- d)4860
Correct answer is option 'A'. Can you explain this answer?
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A radioactive material decays by simultaneous emission of two particle...
when No is initial number of atoms
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A radioactive material decays by simultaneous emission of two particle...
To solve this problem, we need to understand the concept of radioactive decay and how it relates to half-lives.
Radioactive decay is a process where an unstable nucleus of an atom emits particles or radiation in order to become more stable. The rate at which this decay occurs is measured by the half-life of the radioactive material, which is the time it takes for half of the material to decay.
In this problem, we are given that the material decays by simultaneous emission of two particles with respective half-lives of 1620 and 810 years. Let's denote the initial amount of the material as N0.
After the first half-life of 1620 years, half of the material will remain, which is N0/2.
After the second half-life of 1620 years, half of the remaining material will decay, leaving N0/2 * 1/2 = N0/4.
Similarly, after the first half-life of 810 years, half of the material will remain, which is N0/2.
After the second half-life of 810 years, half of the remaining material will decay, leaving N0/2 * 1/2 = N0/4.
Therefore, after a total time of 1620 + 810 = 2430 years, one-fourth of the material remains.
Hence, the correct answer is option 'A' (1080 years).
Radioactive decay is a process where an unstable nucleus of an atom emits particles or radiation in order to become more stable. The rate at which this decay occurs is measured by the half-life of the radioactive material, which is the time it takes for half of the material to decay.
In this problem, we are given that the material decays by simultaneous emission of two particles with respective half-lives of 1620 and 810 years. Let's denote the initial amount of the material as N0.
After the first half-life of 1620 years, half of the material will remain, which is N0/2.
After the second half-life of 1620 years, half of the remaining material will decay, leaving N0/2 * 1/2 = N0/4.
Similarly, after the first half-life of 810 years, half of the material will remain, which is N0/2.
After the second half-life of 810 years, half of the remaining material will decay, leaving N0/2 * 1/2 = N0/4.
Therefore, after a total time of 1620 + 810 = 2430 years, one-fourth of the material remains.
Hence, the correct answer is option 'A' (1080 years).
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A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years.The time, in years, after which one-fourth of the material remains isa)1080b)2430c)3240d)4860Correct answer is option 'A'. Can you explain this answer?
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