Question

The half-life of phosphorus-32 is 14.26 days. calculate its decay constant. express the decay constant numerically in inverse days.

Answer 1

When this is the first order reaction of the radioactive decay:

and according to the first order rate equation:

㏑(Ao/A) = K t

and when Ao = is the starting concentration

and A = is the concentration at time t

and K is the rate constant 

t is the elapsed time = 14.26 days

for 1 half-life ∴ Ao/A = 2

by substitution:

㏑2 = K * 14.26 days

∴ K = 0.0486 days^-1

Answer 2

The decay constant(λ) = 0.0486/day

Further explanation

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

The half-life can be expressed in a decay constant( λ)

$\displaystyle t_ {1/2} = {\dfrac {\ln (2)} {\lambda}}$

The half-life of phosphorus-32 is 14.26 days

Then : t 1/2 = 14.26 days

$\rm \lambda=\dfrac{ln\:2}{14.26}\\\\\lambda=0.0486/day$

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