Question

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to 1 % of its original value ?
a) 5 years
b) 5.66 years
c) 6.65 years
d) 4.60 years​

Answer 1

Answer:

6.65 years

Explanation:

(a) Let the half-life be T years.

original amount of isotope is N

o

​

after decay, the remaining amount is N

now as per given values, N/N

o

​

=3.125%=1/32

also, N/N

o

​

=e

−λt

where λ is decay constant

also λ=0.693/T

so, t=3.4657/λ=5T years.

(b) Let the half-life be T years.

original amount of isotope is N

o

​

after decay, the remaining amount is N

now as per given values, N/N

o

​

=1%=1/100

also, N/N

o

​

=e

−λt

where λ is decay constant

also λ=0.693/T

so, t=4.6052/λ=6.645T years.

Answer 2

Solution :

As per the given data,

• The half-life of the radioactive substance = T yrs
• Let the original amount of radioactive isotope be = N
• Amount of radioactive isotope left = N₀ = 1 N / 100

Now as per the formula,

$\implies\boxed{\sf{\dfrac{N_o}{N} = e^{ - \lambda t }}}$

Now let's substitute the values,

$\implies\sf{\dfrac{N}{100N} = e^{- \lambda t }}$

$\implies\sf{\dfrac{1}{100} = e^{- \lambda t }}$

$\implies\sf{ln( 1) - ln(100) = - \lambda t }$

$\implies\sf{0 - 4.605 = - \lambda t }$

$\implies\sf{ 4.605 = \lambda t }$

We know that,

$\implies\boxed{\sf{ \lambda = \dfrac{0.693}{T}}}$

Hence,

$\implies\sf{4.605 = \dfrac{0.693}{T}\times t}$

$\implies\sf{t = \dfrac{4.605T}{0.693}}$

$\implies\boxed{\sf{t = 6.64T \: yrs}}$

Hence,

It will take 6.64T yrs for the activity to reduce 1% of its original value.