Afreshly prepared sample of curium-243 undergoes 3312 disintegrations per second. after 12.0 yr, the activity of the sample declines to 2292 disintegrations per second. the half-life of curium-243 is years.

Given:
Initial quantity of the radiation = No =  3312 disintegrations per second
Final quantity of radiation = Nt =  2292 disintegrations per second
time = t = 12 years
 
We know that, 
   t(1/2) = t / (㏒1/2 (Nt/ No)
∴ t(1/2) = 12 / (㏒1/2 (2292/3312)
             = 22.59 years.

The half-life of Curium-243 is 22.59 years

22.59 years


Explanation:


1) The law of disintegration decay is expressed as an exponential equation:

N = No x e ^ (-kt)

Where No is the initial number of particles (atoms), k is the disintegration constant, t is the time, and N is the number of atoms remaining after time t has elapsed.


2) By substituting N = 2292, No = 3312, and t = 12 years, you obtain the value of k:


2292 = 3312 x e^(-k×12) => -12k = ln(2292/3312) = -0.6368127 => k = 0.030677


3) Now, you can use the half-time equation (you can derive it from the equation N = No x e ^(-kt) by doing N = No / 2):


t-half = ln(2)/k = ln(2)/0.030677 = 22.59 years

Answer) by studying proton collisions hope this : )

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.