A container is filled with a radioactive substance for which the half-life is 2 days. A week later, when the container is opened, it contains $5 g$ of the substance. Approximately how many grams of the substance were initially placed in the container?

40

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

80

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $60$
initially placed amount $N_{0}$ and present be $N$ then we have
$N_{0} / N = e^{0.693 t / t_{1 / 2}}$
put $t = 7$ and $t_{1 / 2} = 2$ and $N = 5 g m$ we get $N_{0} = 11.3 \times 5 = 56.5 g m$

Suggest Corrections

Similar questions

A radioactive substance of half life 69.3 days is kept in a container. The time in which 80 % of the substance will disintegrate will be

A radioactive substance of half life 69.3 days is kept in a container. The time in which 80% of the substance will disintegrate will be [take ln(5) = 1.61]

Q. A radioactive substance decays from

100

grams to

6.25

grams in

100

days. What is the half-life of this radioactive substance?

Q. A radioactive substance contains

10000

nuclei and its half-life period is

20

days. The number of nuclei present at the end of

10

days is

Q. A radioactive substance has half-life

5760 y r s

. A sample of

100 m g

containing substance is reduced to

25 m g

in:

Join BYJU'S Learning Program

A container is filled with a radioactive substance for which the half-life is 2 days. A week later, when the container is opened, it contains 5g of the substance. Approximately how many grams of the substance were initially placed in the container?

The correct option is B 60initially placed amount N0 and present be N then we have N0/N=e0.693t/t1/2put t=7and t1/2=2 and N=5gm we get N0=11.3×5=56.5gm

A container is filled with a radioactive substance for which the half-life is 2 days. A week later, when the container is opened, it contains $5 g$ of the substance. Approximately how many grams of the substance were initially placed in the container?

The correct option is B $60$
initially placed amount $N_{0}$ and present be $N$ then we have
$N_{0} / N = e^{0.693 t / t_{1 / 2}}$
put $t = 7$ and $t_{1 / 2} = 2$ and $N = 5 g m$ we get $N_{0} = 11.3 \times 5 = 56.5 g m$