if the number of radioactive atoms at time t is 2x10^6, and 2x10^4 atoms disintegrate in 5 min, what is the approximate radioactive constant? show steps.
Answer should be 2*10^-3
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if the number of radioactive atoms at time t is 2x10^6, and 2x10^4 atoms disintegrate in...
Two samples of a radioactive materal each contain No atoms at time t = 0. At later times the number of undecayed atoms in sample A is: Na = Noe^-20t and for sample B is: Nb = Noe^-10t Which sample decays more rapidly? Explain your answer.
The half-life of a radioactive isotope is 2.5 min. If there are initially 48 Times 10^32 atoms of this isotope, (a) the number of atoms of this isotope remaining after 15 min is: 12 Times 10^32 7.5 Times 10^31 3 Times 10^32 6 Times 10^4 3 Times 10^2 (b) what is the decay constant? (c) What is the initial activity?
The number of radioactive isotopes in a sample, N(t) as a
function of time is given by an exponential law
where N(0) is the initial number of radioactive
isotopes at time t=0, and k is a constant. Find the expression for
t1/f, the time it takes for N(t) to go from its initial
value to N(o)/f. What is the value for X in the following
expression?
If a radioactive isotope has half-life h, then the number of atoms in a sample is given by a(t)=a-2-/h where ao gives the number of atoms in the sample at time 0. The radio active decay of carbon-14 leads to clever way of determining the age of fossils and remnants of plants and animals. When an organism dies, the radioactive carbon- 14 in it decays, with half of it gone in 5,730 years. a. Suppose a mummified cat has a...
This is all the information given. There is nothing about
half-lives or the initial number of elements.
Question 1 0/0.83 pts If 434 of the atoms of a radioactive isotope disintegrate each day, what is the decay constant of the process? Round answer to 4 decimal places You Answered 0.005 Correct Answer 0.0023
I only need help with part c). Thanks!
As a radioactive specimen decays, its activity decreases exponentially as the number of radioactive atoms diminishes. Some radioactive species have mean lives in the millions (or even billions) of years, so their exponential decay is not readily apparent. On the other hand, many species have mean lives of minutes or hours, for their exponential decay is easily observed. According to the exponential decay law, the number of radioactive atoms that remain after...
could you do and explain part a
er counts the number of decays from a radioactive sample ina e interval Δt from a radioactive source, starting at time t 0, The limiting n for this kind of experiment is the exponential distribution (5.69) wthere T is a positive constant. (a) Sketch this function. The distribution is zero for ent begins only at 0.) (b) Prove that this function satisfies the normalization condition (5.13). () Find the mean time T at...
Radioactive Decay - Half-life and Activity 1 Radioactive decay - Half-life Time 0 1000 21 31 750 N 1.000.000 500,000 250,000 125,000 62,500 31.250 15.625 7813 3.506 1.953 977 51 6 500 7 BI . 101 250 125 0 tie 21.234.41516171819, 1012 Time in multiples of A radioactive sample's half-life is 30.2 years. 1 year = 365 days, 1 day = 24 hours, 1 hour - 60 minutes, 1 minute = 60 seconds (a) Find its decay constant in year...
help please!!
yeah sorry here is an updated page
I
need the In daughter atoms i don't know how to the calculations for
it
7. Scientists like to use data to build physical models (or mathematical models to predict how data will look). Linear trends are easiest to work with; so we often look for ways to identify linear trends in a data set. Nuclear reactions all follow first order kinetics meaning the rate of a nuclear reaction is given...
3 pts It is handy to have an equation to quickly determine the number of atoms left in a radioactive sample as a function of time. For this we can divide the initial amount by two for every half-life of time, the following equation does exactly that: N No (24/01/2) N, 2 1/2 where N is the initial number of atoms at t=0, t is time passed and is the half-life. Use the above equation to help you answer the...