A new element, Transium, was discovered. It has a decay constant of 2.91 × 10^-3
ID: 1517252 • Letter: A
Question
A new element, Transium, was discovered. It has a decay constant of 2.91 × 10^-3 s^-1 and is a known alpha-particle emitter. Part A: At a certain time, 93.1 g of Transium are present. How much remains (Nf) after 8.65 minutes? Part B: What happens to the atomic number of one of the decaying nuclei? 1) The atomic number of the daughter nucleus is the same as the atomic number of the Transium. 2)The atomic number of the daughter nucleus is greater than the atomic number of the Transium. 3)The atomic number of the daughter nucleus is less than the atomic number of the Transium. 4)There is not enough information to answer the question.
Explanation / Answer
given
lamda = 2.91*10^-3 s^-1
mo = 93.1 grams
A) at time t = 8.65 min
= 8.65*60
= 519 s
Apply, m = mo*e^(-lamda*t)
= 93.1*e^(-2.91*10^-3*519)
= 20.56 grams <<<<<<---------------Answer
B)
3)The atomic number of the daughter nucleus is less than the atomic number of the Transium.
the atomic number decreases by 2. because alfa particle consists of 2 protons and 2 neutrons.
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