## Question

### Solution

Correct option is

40 seconds

The 6.25% of the reduced number (12.5%) is

Suppose, after 10 seconds, the number of undecayed nuclei reduces from 12.5% to 0.78125% in an interval of n half-lives. Then

or        n = 4.

= 40 seconds.

#### SIMILAR QUESTIONS

Q1

The activity of a radioactive element reduces to th of its original value in 30 years. Find the half-life and the decay constant of the element.

Q2

The half-life of thorium-X is 3.64 days. After how many days will 0.1 of the mass of a sample of the substance remain undecayed?

Q3

How much of 5.00 gram of polonium will decay in one year? The half-life of polonium is 138 days.

Q4

The half-life of a radioactive substance is  against . Calculate the decay rate for  atoms of the substance.

Q5

The half-life of  against . Calculate the activity of 1 g sample of . Avogadro’s number is .

Q6

The half-life of  is 28 years. How much is the disintegration rate of 15 mg of this isotope? The Avogadro’s number is .

Q7

1 g of particles per second. Find its half-life and average life. Avogadro’s number is .

Q8

Determine the amount of  required to provide a source of -particles of 5 millicurie strength. The half-life of Po is 138 days. The Avogadro’s number is

Q9

The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. 5 minute after, the rate becomes 2700 disintegrations per minute. Calculate the half-life of the sample. .

Q10

The normal activity of a living mater containing radioactive carbon C14 is found to be 15 decays per minute per gram of carbon. An archaeological specimen shows an activity of 9 decays per minute per gram of carbon. Estimate the age of the specimen. The half-life of C14 is 5730 years.