The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6/25% of its initial value. The value of x is___________.

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

According to Nuclear activity, we can write

$N_{0} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{2} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{4} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{8} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{16} = (0.0625) N_{0}$

Time required = 4 × $t_{\frac{1}{2}} = 20 y r s$

0 Upvotes 0 Downvotes

Similar Questions for you

The nucleus 115/48cd, after two successive β?decay will give

Raj Pandey

Two successive β decays increase the charge no. by 2.

A radioactive sample disintegrates via two independent decay processes having half lives T?/??¹? and T?/??²? respectively. The effective half-life, T?/? of the nuclei is:

Raj Pandey

From Radioactive Decay Law,
dN/dt = λ? N + λ? N = λ_eff N
⇒ λ_eff = λ? + λ? ⇒ ln (2)/T = ln (2)/T? + ln (2)/T? ⇒ T = (T? ) / (T? + T? )
(where T, T? , and T? are half-lives)

In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is

alok kumar singh

The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.

The activity of a radioactive sample falls from $700 s^{- 1}$ to $500 s^{- 1}$ in 30 minutes. Its half life is close to

alok kumar singh

$A = A_{0} e^{- λ t}$

$\begin{array}{r} 500 = 700 e^{- λ t} \\ λ t = l n ? \frac{7}{5} \\ \frac{l n ? 2}{t_{1 / 2}} \times 30 = l n ? \frac{7}{5} \\ ? t_{1 / 2} = \frac{l n ? 2 \times 30}{l n ? \frac{7}{5}} \end{array}$

$t_{1 / 2} = 61.8 m i n \approx 62 m i n$ .

The half-life of ¹??Au is 3 days. If atomic weight of ¹??Au is 198 g/mol then the activity of 2 mg of ¹??Au is [ in disintegration / second ] :

Raj Pandey

t? /? = 3 days = 72 hours
dN/dt = λN = (ln2/t? /? ) N
= (0.693 × 6.02×10²³ × 2×10? ³) / (72 × 3600 × 198)
= 1.618 × 10¹³
= 16.18 × 10¹² disintegration/second

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6/25% of its initial value. The value of x is___________.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question