Home
Class 12
PHYSICS
If a radioactive substance reduces to (1...

If a radioactive substance reduces to `(1)/(16)` of its original mass in `40` days, what is its half-life ?

A

10 days

B

20 days

C

40 days

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
From the formula,
`N= N_(0)(1/2)^(n)`
`N/16 = N_(0)(1/2)^(n)`
`N_(0)` = original numbe of atom
`(1/2)^(4) = (1/2)^(n)`
`rArr n=4`
4 half-lives
`therefore 4T_(1//2) = 40`
`T_(1//2) = 40/4 = 10` days
Promotional Banner

Topper's Solved these Questions

  • NULECUS

    BITSAT GUIDE|Exercise BITSAT Archives|8 Videos

  • MAGNETIC FIELD

    BITSAT GUIDE|Exercise All Questions|52 Videos

  • PHYSICS FOR GASEOUS STATE

    BITSAT GUIDE|Exercise BITSAT Archives|8 Videos

Similar Questions

Explore conceptually related problems

The activity of a radioactive element reduces to (1//16)th of its original value in 30 years. Find its half life?

A radioactive substance decay to 1/5 of its original value in 56 days. The decay constant is

A radioactive substance reduces to 1/32 of its original value in 300 days. The half life of radioactive substance is

A radioactive substance decays to ((1)/(16))^(m) of its initial activity in 40 days. The half-life of the radioacctive substance expressed in days is

A radioactive substance decays to ((1)/(16))^(th) of its initial activity in 80 days. The half life of the radioactive substance expressed in days is …….

Ceratain radioactive substance reduces to 25% of its value is 16 days . Its half-life is

A radioactive substance decays to ( 1/10)^th of its original value in 56 days. Calculate its decay constant.

A certain substance decays to 1//32 of its initial activity in 25 days. Calculate its half-life.

After 2 hours the amount of a certain radioactive substance reduces to 1//16^(th) of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is