Question

half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes initially the examples of A and B have equal number of nuclei after 80 minutes the ratio of remaining number of A and B nuclei is ​

Answer 1

Nᴀ/Nʙ =1/4

Explanation:

total time given = 80 min

•Number of half life's of A,

nA = 80 min/20min = 4

•Number of half life of B, nB = 80 min/40 min = 2

•Number of nuclei remained under DK undecayed

N = No = [ 1/2 ]^n

Where No is initial number of nuclei and N is final number of nuclei

So for two different cases (A) and (B),

Nᴀ/Nʙ = [ 1/2 ]^nA / [ 1/2 ]^nB

Nᴀ/Nʙ = [ 1/2 ]^4 / [ 1/2 ]

Nᴀ/Nʙ = [ 1/16 ] / [ 1/4 ]

Nᴀ/Nʙ = 1 /4

The Ratio of remaining number of

A and B nuclei is 1/4 !

Answer 2

Answer:

$\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{\red{ Answer:}}}}}}}}\end{gathered}$

Nᴀ/Nʙ =1/4

Explanation:

total time given = 80 min

•Number of half life's of A,

nA = 80 min/20min = 4

•Number of half life of B, nB = 80 min/40 min = 2

•Number of nuclei remained under DK undecayed

N = No = [ 1/2 ]^n

Nᴀ/Nʙ =1/4

Explanation:

total time given = 80 min

•Number of half life's of A,

nA = 80 min/20min = 4

•Number of half life of B, nB = 80 min/40 min = 2

•Number of nuclei remained under DK undecayed

N = No = [ 1/2 ]^n

Where No is initial number of nuclei and N is final number of nuclei

So for two different cases (A) and (B),

Nᴀ/Nʙ = [ 1/2 ]^nA / [ 1/2 ]^nB

Nᴀ/Nʙ = [ 1/2 ]^4 / [ 1/2 ]

Nᴀ/Nʙ = [ 1/16 ] / [ 1/4 ]

Nᴀ/Nʙ = 1 /4

The Ratio of remaining number of

A and B nuclei is 1/4 !

Where No is initial number of nuclei and N is final number of nuclei

So for two different cases (A) and (B),

Nᴀ/Nʙ = [ 1/2 ]^nA / [ 1/2 ]^nB

Nᴀ/Nʙ = [ 1/2 ]^4 / [ 1/2 ]

Nᴀ/Nʙ = [ 1/16 ] / [ 1/4 ]

Nᴀ/Nʙ = 1 /4

The Ratio of remaining number of

A and B nuclei is 1/4 !