Physics

Radioactivity and Nuclear Decay

74 Questions

Radioactivity and nuclear decay problems revolve around half-life calculations, decay rates, and nuclear stability. These topics are crucial for general science preparation in many exams. Regular practice helps in quickly solving complex decay chain numericals.

half life calculationnuclear stabilitydecay rate constantalpha beta decay

Radioactivity and Nuclear Decay Questions

Multiple choice

The isotope used for determining the age of water is

  1. chlorine-36

  2. magnesium-27

  3. lead-210 beta 22

  4. potassium-40

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Chlorine-36 emitts beta particles.Its half life is 301,000 years.It is used in the measurement of sources of chloride and determining the age of water up to about 2 million years old.It is a naturally occurring radioisotope .

Multiple choice

The radioactive nuclide used in geological dating is

  1. Y-90

  2. U-238

  3. B-10

  4. P-32
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Geological dating covers thousands of years. Only isotopes with long half-lives can be used. The isotope with the longest half-life is uranium-238. Uranium-238 can be used to date rocks, fossils and meteorites. Carbon-14 is used to date matter that was once alive.

Multiple choice

The half-life of a radioactive substance is 30 minutes. During 2 hours, the fraction of the total number of atoms that have decayed would be ____________.

  1. 6.25%

  2. 12.5%

  3. 25%

  4. 87.5%

  5. 92.75%
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Half-life = 30 minutes For 2 hrs, the number of half-life (n) = 4 Cn = 1/2n= 1/24 = 1/16 = 6.25% So, after 2 hrs the amount of radioactive substance will be reduced to 1/16th (6.25%) of its initial amount. Hence, the amount decayed during 2 hrs = (100 - 6.25)% = 92.75%

Multiple choice

Atomic hydrogen has a life period of

  1. one minute

  2. one hour

  3. fraction of second

  4. one day
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Atomic Hydrogen (H.) has only one electron which makes it highly unstable. Whereas molecular Hydrogen has covalent bond which includes sharing of electrons between two hydrogen atoms. This becomes very stable condition like He atom. Thats why atomic hydrogen has very small life period.

Multiple choice

Half life period of C-14 is

  1. 5730 years

  2. 15240 years

  3. 8190 years

  4. 8208 years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radioactive decay of carbon-14 is exponential with half-life of 5,730 years. A quantity of carbon-14 will decay to half of its original amount (on average) after 5,730 years, regardless of how big or small the original quantity was. 

Multiple choice chemistry nuclear physics types of radioactivity nuclear chemistry and radioactivity radioactivity

The count rate observed from a radioactive source at $t$ second was $N _0$ and at $4t$ second it was $\dfrac{N _0}{16}$. The count rate observed at $\left(\dfrac{11}{2}\right)t$ second will be

  1. $\dfrac{N _0}{128}$
  2. $\dfrac{N _0}{64}$
  3. $\dfrac{N _0}{32}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
after n lives,

$N=N _0\left ( \dfrac{1}{2} \right )$

initially,

$t=t$

$t=4t-t=3t$

$\dfrac{N _0}{16}=N _0\left ( \dfrac{1}{2} \right )^n$

$n=4$

$4t _{\frac{1}{2}}=3t$

$\Rightarrow t _{\frac{1}{2}}=\dfrac{3}{4}t$

$\lambda =\dfrac{\ln 2}{t _{\frac{1}{2}}}=\dfrac{4\ln 2}{3t}$

at $t=\dfrac{11}{2}ts$

$t=\dfrac{11}{2}t-t=\dfrac{9}{2}t$

$N=N _0e^{-\lambda \frac{9}{2}t}$

$N=N _0e^{-\lambda \frac{4\ln 2}{3t}\times \frac{9}{2}t}$

$N=N _0e^{-6\ln 2}$

$N=N _0e^{\ln \frac{1}{2^6}}$

$N=\dfrac {N _0}{2^6}=\dfrac {N _0}{64}$
Multiple choice chemistry nuclear physics types of radioactivity nuclear chemistry and radioactivity radioactivity

After $280$ days, the activity of a radioactive sample is $6000 dps$. The activity reduces to $3000 dps$ after another $140 days$. The initial activity of the sample in $dps$ is:

  1. $6000$
  2. $9000$
  3. $3000$
  4. $24000$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, the activity of the radioactive sample reduces to half in $140$ days. Therefore, the half life of the sample is $140$ days. $280$ days is it's two half lives. So before two half lives it's activity was

${2^6} \times 6000 = 24000\,dps.$
Hence, the option $D$ is the correct answer.

Multiple choice chemistry nuclear physics types of radioactivity nuclear chemistry and radioactivity radioactivity

The radioactivity of an old sample of a liquid due to tritium (half life $12.5$ years) was found to be only about $3$% of that measured in a recently purchased bottle marked $7$ year old. The sample must have been prepared about:

  1. $70$ year
  2. $63.24$ year
  3. $420$ year
  4. $300$ year
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Hence, \ \frac { N }{ { { N _{ 0 } } } } =0.03=\left( { -\lambda t } \right)  \ In\left( { 0.03 } \right) =-\lambda t=\frac { { -In\left( 2 \right) \times t } }{ { \frac { { { T _{ 1 } } } }{ 2 }  } }  \ t=\frac { { In\left( { 0.03 } \right) \times \frac { { { T _{ 1 } } } }{ 2 }  } }{ { -In\left( 2 \right)  } } =63.24\, years \end{array}$

Hence,
option $(B)$ is correct answer.

Multiple choice chemistry nuclear physics types of radioactivity nuclear chemistry and radioactivity radioactivity

The radioactivity of a sample is $R _ { 1 } $at a time $T _ { 1 }$ and $R _ 2$ at a time $ T _ 2 $ . If the half-life of the specimen is T  ,  the number of atoms that have disintegrated in the time $T _ 2 - T _  1 $ is proportional to 

  1. $T _2-T _1\alpha(R _1-R _2)T$
  2. $\left( R _ { 1 } - R _ { 2 } \right)$
  3. $\left( R _ { 1 } - R _ { 2 } \right) / T$
  4. $\left( R _ { 1 } - R _ { 2 } \right) T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
[D]
$\begin{array}{l} { R _{ 1 } }=\lambda { N _{ 1 } } \\ { R _{ 2 } }=\lambda { N _{ 2 } } \end{array}$
Number of atoms disintegrated in time 
$\begin{array}{l} { T _{ 2 } }-{ T _{ 1 } }={ N _{ 1 } }-{ N _{ 2 } }=\dfrac { { { R _{ 1 } }-{ R _{ 2 } } } }{ \lambda  }  \\ \lambda =\dfrac { { 0.693 } }{ T }  \\ { T _{ 2 } }-{ T _{ 1 } }=\dfrac { { { R _{ 1 } }-{ R _{ 2 } }\times T } }{ { 0.693 } }  \\ { T _{ 2 } }-{ T _{ 1 } }\alpha \left( { { R _{ 1 } }-{ R _{ 2 } } } \right) T \end{array}$