Suppose you start with a sample with 6.030×10^8 nuclei of a particular isotope. this isotope has a half-life of 537 s. what is the decay constant for this particular isotope? how many nuclei remain after 1.50×10^3 s?
what is the activity of the sample at the beginning of the experiment?

decay constant is 1.290  ×  s-1

Number is  0.870  ×  nuclei

activity is 1.1223 × Ci

Explanation:

Given data

isotope N1 = 6.030×10^8 nuclei

half life  T/2 = 537 s

to find out

decay constant and How many nuclei remain and activity of the sample at the beginning

solution

we know  that half life  and decay constant relationship that is

N = N1 ×   ...............1

here  k is decay constant and we know dN / dt is equal to -k × N

and T/2 = 0.693/k

so k = 0.693 / (T/12)

put here T/2

k = 0.693 / 537 = 1.290  ×  s-1

so decay constant is 1.290  ×  s-1

and

by equation 1 we get

N = N1 ×  

N = 6.030×10^8 ×  1.50×10^3}" alt="10^{-3}" />  1.50×10^3}" />  

N = 0.870  ×  nuclei

and

we know Activity of sample  =k×N

so

activity = 0.870 ×  × 1.290  ×

activity = 1.1223 × Ci

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

object 2 because all the other ones are wrong