datavorous/entrance-exam-dataset huggingface dataset:
A radioactive material decays by simultaneous emission of two particles with half-lives 1620 yr and 810 yr respectively. The time (in yr) after which one-fourth of the material remains, is

Options:
A 4860 yr
B 3240 yr
C 2340 yr
D 1080 yr

The correct answer is:
1080 yr

Solution:
From Rutherford-Soddy law, the number of atoms left after n half-lives is given by
N=N012n
Where, N0 is the original number of atoms.
The number of half-life
n=time of decayeffective half-life
Relation between effective disintegration constant (Î») and half-life (T) is
Î»=lnâ¡2T
â´ Î»1+Î»2=lnâ¡2T1+lnâ¡2T2
Effective half -life
1T=1T1+1T2=11620+1810
1T=1+21620âTâ540 yr
â´ n= t540
â t540=2
â t= 2 Ã 540 = 1080 yr

Topic: Physics, Nuclear Physics, NEET
A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years, respectively. What will be the time after the which one third of the material remains ? (Take ln3=1.1)

Options:
A 1110 years
B 700 years
C 340 years
D 740 years

The correct answer is:
740 years

Solution:
Given λ1=ℓn2700/year λ2=ℓn21400/year
∴λnet =λ1+λ2=ln21700+11400
=3ln21400/year
Now, let initial number of radioactive nuclei be
N0,
∴N03=N0e-λmatt

⇒ℓn13=-λnett
⇒1.1=3×0.6931400t⇒t≈740 years
Hence, option 4.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A radioactive material has a half-life of 10 days. What fraction of the material would remain after 30 days?

Options:
A 0.5
B 0.25
C 0.125
D 0.33

The correct answer is:
0.125

Solution:
$T_{1 / 2}=10$ days, $T=30$ days
$$
\begin{aligned}
& T=n \times T_{1 / 2} . \Rightarrow n=\frac{T}{T_{1 / 2}}=\frac{30}{10}=3 \\
& \frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^3=\frac{1}{8}=0.125 .
\end{aligned}
$$

Topic: Physics, Nuclear Physics, AIIMS, AIIMS 2005
A radioactive material has mean lives of 1620 years and 660 years for α and β emissions respectively. The material decays by simultaneous α and β emission. The time in which 14th of the material remains intact is -

Options:
A 4675 years
B 720 years
C 650 years
D 324 years

The correct answer is:
650 years

Solution:
Effective mean life for parallel disintegration is given by
Ï=Ï1Ï2Ï1+Ï2=1620Ã6602280= 469 year
Let t be the time in which 1/4th of the material remains intact
N4=N2tÏ
t=2Ï ln2
t=2Ã469Ã0.693
t = 650 year

Topic: Physics, Nuclear Physics, NEET
A radio-active material is reduced to $1 / 8$ of its original amount in 3 days. If $8 \times 10^{-3} \mathrm{~kg}$ of the material is left after 5 days. The initial amount of the material is

Options:
A $700 \mathrm{gm}$
B $900 \mathrm{gm}$
C $475 \mathrm{gm}$
D $256 \mathrm{gm}$

The correct answer is:
$256 \mathrm{gm}$

Solution:
$\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\mathrm{N}=\frac{\mathrm{N}_0}{8}$
$\frac{\mathrm{N}_0}{8}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\mathrm{n}=3$
3 half lives $=3$ days
1 half life $=1$ day
5 days $=5$ half life
$\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 8 \times 10^{-3}=\mathrm{N}_0\left(\frac{1}{2}\right)^5$
$\Rightarrow \quad N_0=2^5 \times 8 \times 10^{-3}=256 \mathrm{gm}$

Topic: Physics, Nuclear Physics, JEE Main
A radioactive material whose half life period is 2 years weighs $1 \mathrm{~g}$ and is stored in the laboratory for 4 years. Then the amount of remaining radioactive material is

Options:
A $0.5 \mathrm{~g}$
B $0.125 \mathrm{~g}$
C $0.25 \mathrm{~g}$
D $0.0625 \mathrm{~g}$

The correct answer is:
$0.25 \mathrm{~g}$

Solution:
Half life, $\mathrm{T}_{1 / 2}=2$ year
Quantity of substance, $\mathrm{N}_0=1 \mathrm{~g}$
Using law of radio activity, $\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$
Decay constant, $\lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}$
$\begin{aligned}
& =\frac{0.693}{2}=0.346 \\
& =0.35 \text { per year } \\
& \text { Thus, } \mathrm{N}=1 \times \mathrm{e}^{-0.35 \times 4}=0.25 \mathrm{~g}
\end{aligned}$

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
A radioactive nuclei with decay constant $0.5 / \mathrm{s}$ is being produced at a constant rate of 100 nuclei/s. If at $t=0$ there were no nuclei, the time when there are 50 nuclei is:

Options:
A
$1 \mathrm{~s}$

B
$2 \ln \left(\frac{4}{3}\right) \mathrm{s}$

C
$\ln 2 \mathrm{~s}$

D
$\ln \left(\frac{4}{3}\right) \mathrm{s}$

The correct answer is:
$2 \ln \left(\frac{4}{3}\right) \mathrm{s}$

Solution:
Let $N$ be the number of nuclei at any time $t$ then,
$$
\begin{aligned}
&\frac{d N}{d t}=100-\lambda N \\
&\text { or } \int_0^N \frac{d N}{(100-\lambda N)}=\int_0^t d t \\
&-\frac{1}{\lambda}[\log (100-\lambda N)]_0^N=t \\
&\log (100-\lambda N)-\log 100=-\lambda t \\
&\log \frac{100-\lambda N}{100}=-\lambda t \\
&\frac{100-\lambda N}{100}=e^{-\lambda t} \\
&1-\frac{\lambda N}{100}=e^{-\lambda t}
\end{aligned}
$$
$$
\begin{aligned}
&N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right) \\
&\text { As, } N=50 \text { and } \lambda=0.5 / \mathrm{sec} \\
&\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5 t}\right)
\end{aligned}
$$
Solving we get,
$$
t=2 \ln \left(\frac{4}{3}\right) \mathrm{sec}
$$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2014 (11 Apr Online)
A radioactive nucleus \( A \) has a single decay mode with half life \( \tau_{\mathrm{A}} \). Another radioactive nucleus \( B \) has two decay modes, \( 1 \) and \( 2 \). If decay mode \( 2 \) is absent, the half life of \( B \) would be \( \frac{\tau_{\mathrm{A}}}{2} \). If decay mode \( 1 \) is absent, the half life of \( B \) would be \( 3 \tau_{\mathrm{A}} \) . If the actual half life of \( B \) is \( \tau_{\mathrm{B}} \), then the ratio \( \frac{\tau_{\mathrm{B}}}{\tau_{\mathrm{A}}} \) is,

Options:
A \( \frac{3}{7} \)
B \( \frac{7}{2} \)
C \( \frac{7}{3} \)
D \( 1 \)

The correct answer is:
\( \frac{3}{7} \)

Solution:
This is an example of parallel decay of a nuclide.

For this decay, we can write,

-dNBdt=dNCdt+dNDdt⇒N0λB=N0λ1+N0λ2 λ=deecay constant⇒λB=λ1+λ2

Decay constant is related to half life as, λ=ln2τ.

⇒ln2τB=ln2τ1+ln2τ2, where τ1 and τ2 are half life for paths 1 and 2, respectively.

⇒τB=τ1×τ2τ1+τ2=τA2×3τAτA2+3τA=37τA⇒τBτA=37

Topic: Physics, Nuclear Physics, JEE Main
A radioactive nucleus A with a half-life T, decays into a nucleus B. At t=0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by: Consider logex=logx

Options:
A t=Tlog⁡1.3
B t=T2logâ¡Â 2logâ¡Â 1.3
C t=Tlogâ¡ 1.3logâ¡ 2
D t=Tlog1.3

The correct answer is:
t=Tlogâ¡ 1.3logâ¡ 2

Solution:
Let N A and N B be the number of molecules of A and B after time t.
Also, after time t, NBNA=0.3(given).

Also, let N 0 be the total number of nucleus initial
After time t, N A + N B = N 0
N A +0.3 N A = N 0
∴   N A = N 0 1.3
Also, the rate of disintegration of A
N A = N 0 e −λt
⇒ N 0 1.3 = N 0 e −λt ;  1 1.3 = e −λt    or  log( 1.3 )=λt
t=log(1.3)λ

∴t=Tlog(1.3)log(2)=Tlog(1.3)log2.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2017 (02 Apr)
A radioactive nucleus AZX undergoes spontaneous decay in the sequence AZX→Z-1B→Z-3C→Z-2D, where Z is the atomic number of element X. The possible decay particles in the sequence are:

Options:
A α,β-,β+
B α,β+,β-
C β+,α,β-
D β-,α,β+

The correct answer is:
β+,α,β-

Solution:
As we know due to α-decay atomic number decreases by 2 and mass no. decreases by 4 . And due to β-decay mass no. remain unchanged and atomic no. increases by 1 And due to β+decay mass no. remain unchanged but atomic no. decreases by 1.

So, option (3) is correct option.

Topic: Physics, Nuclear Physics, NEET, NEET 2021
A radioactive nucleus at rest having mass number A , emits an alpha-particle with speed v. The recoil speed of the daughter nucleus is

Options:
A 2v(A-4)
B 2v(A+4)
C 4v(A-4)
D 4v(A+4)

The correct answer is:
4v(A-4)

Solution:
As per the law of conservation of momentum,
0=4v-A-4vr
â´ Recoil speed vr=4v(A-4)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective halflife of the nucleus will be

Options:
A 3.75 hours
B 0.56 hours
C 0.26 hours
D 1.80 hours

The correct answer is:
1.80 hours

Solution:
Here the particle is decayed by two processes with decay constant λ1 and λ2.

Radioactive decay rate is directly proportional to number of nucleus present at any time, so total rate of decay is

dAdt--λ1A+-λ2A

⇒dAdt=-λ1+λ2A

⇒λeff =λ1+λ2

⇒ln2t12eff=ln2t121+ln2t122

⇒t12off =4.5×37.5 hours =1.8 hours

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A radioactive nucleus can decay by two different processes. The half lives of the first and second decay processes are $5 \times 10^3$ and $10^5$ years respectively. Then, the effective half-life of the nucleus is

Options:
A $105 \times 10^5 \mathrm{yr}$
B $4762 \mathrm{yr}$
C $10^4 \mathrm{yr}$
D $47.6 \mathrm{yr}$

The correct answer is:
$4762 \mathrm{yr}$

Solution:
The effective half-life of the nucleus
$\begin{aligned}
\lambda & =\lambda_1+\lambda_2 \\
\frac{0.69}{T} & =\frac{0.69}{T_1}+\frac{0.69}{T_2} \\
T & =\frac{T_1 T_2}{T_1+T_2} \\
& =\frac{5 \times 10^3 \times 10^5}{5 \times 10^3[1+20]}=\frac{5 \times 10^5}{105} \\
& =4762 \mathrm{yr}
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
A radioactive nucleus can decay in two different processes with half life $0.7 \mathrm{hr}$ and $0.3 \mathrm{hr}$.
The effective average life of the nucleus in minutes approximately is (value of $\ln 2=0.7$ )

Options:
A 14
B 18
C 24
D 26

The correct answer is:
18

Solution:
Key Idea If a radioactive sample has more than one radioactive material then the effective half life.
$\frac{1}{T_{\text {eff T/2 }}}=\frac{1}{T_{1 \frac{1}{2}}}+\frac{1}{T_{2 \frac{1}{2}}}+\frac{1}{T_{3 \frac{1}{3}}}+\ldots$
Where, $T_{n \frac{1}{2}}=$ half life of $n^{\text {th }}$ material.
Here, effective half life of the net material,
$\begin{aligned}
& T_{\text {eff } \frac{1}{2}}=\frac{0.7 \times 0.3}{0.7+0.3}=0.21 \mathrm{hr} \\
&\left(\because T_{1 \frac{1}{2}}=0.7 \mathrm{hr}, \mathrm{T}_{2 \frac{1}{2}}=0.3 \mathrm{hr}\right)
\end{aligned}$
As we know, $T_{1 / 2}=0.693 \tau_{\text {aveg. mean life }}$
$\begin{aligned}
\Rightarrow \quad \tau_{\text {aveg. mean life }} & =\frac{T_{\text {eff } \frac{1}{2}}}{0.693} \\
& =\frac{0.21}{0.693}=0.3 \mathrm{hr} . \\
\Rightarrow \quad \tau_{\text {aveg. mean life }} & =0.3 \times 60=18 \mathrm{~min} .
\end{aligned}$
Hence, the correct option is (b).

Topic: Physics, Nuclear Physics, JEE Main
A radioactive nucleus decays by two different processes. The half-life for the first process is 10 s and that for the second is 100 s. The effective half-life of the nucleus is close to:

Options:
A 9 s
B 6 s
C 55 s
D 12 s

The correct answer is:
9 s

Solution:
Using the equation of radioactive decay law,

-dNdt=λ1N+λ2N ...(1)

-dNdt=λ1+λ2N ...(2)

Where,

-dNdt is the rate of decay

λ1, λ2 are decay constant for two different processes

N is the number of atoms.

Now applying the decay law for the combine process, we get

-dNdt=λeffN ...(3)

Equating equation (1) and (2) we get

λeff=λ1+λ2 ...(4)

Now the half life of the radioactive nucleus is given by,

λ=ln2T ...(5)

Where, T is the half life of the radioactive nucleus

Using equation (5) equation (4) becomes,

ln2T=ln2T1+ln2T2

1T=1T1+1T2

1T=110+1100=11100

⇒T=10011≈9 s

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A radioactive nucleus emits $4 \alpha$ particles and $7 \beta$ particles in succession. The ratio of number of neutrons to that of protons is $[\mathrm{A}=$ mass number, $\mathrm{Z}=$ atomic number $]$

Options:
A $\frac{\mathrm{A}-\mathrm{Z}-13}{\mathrm{Z}-2}$
B $\frac{A-Z-13}{Z-1}$
C $\frac{A-Z-15}{Z-1}$
D $\frac{A-Z-11}{Z-2}$

The correct answer is:
$\frac{A-Z-15}{Z-1}$

Solution:
Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$.
When an a-particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4 . So, for given case, $\mathrm{Z} \mathrm{X}^{\mathrm{A}} \stackrel{4 \text { a-particle }}{\longrightarrow} \mathrm{Z}_{-\mathrm{a}} \mathrm{Y}^{\mathrm{A}-16}$
When a $\beta$-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case,
$\mathrm{Z}-8 \mathrm{Y}^{\mathrm{A}-16} \stackrel{7 \beta-p a r t i c l e}{\longrightarrow} \mathrm{Z}-1 \mathrm{Z}^{\mathrm{A}-16}$

Topic: Physics, Nuclear Physics, MHT CET, MHT CET 2020 (16 Oct Shift 2)
A radioactive nucleus (initial mass number A and atomic number Z) emits 3 Î±-particles and 2 positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be

Options:
A A-Z-8Z-4
B A-Z-4Z-8
C A-Z-12Z-4
D A-Z-4Z-2

The correct answer is:
A-Z-4Z-8

Solution:
In positive beta decay a proton is transformed into a neutron and a positron is emitted.
p+→n0 + e+
Number of neutrons initially was A-Z.
Number of neutrons after decay (A-Z) -3×2(due to alpha particles) + 2×1(due to positive beta decay).
The number of protons will reduce by 8 [as 3×2 (due to alpha particles) + 2(due to positive beta decay)].
Hence, atomic number reduces by 8.
So, the ratio number of neutrons to that of protons
=A-Z-4Z-8

Topic: Physics, Nuclear Physics, JEE Main
A radioactive nucleus (initial mass number $A$ and atomic number Z) emits $3 \alpha$-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Options:
A
$\frac{A-Z-8}{Z-4}$

B
$\frac{A-Z-4}{Z-8}$

C
$\frac{A-Z-12}{Z-4}$

D
$\frac{A-Z-4}{Z-2}$

The correct answer is:
$\frac{A-Z-4}{Z-8}$

Solution:
In positive beta decay a proton is transformed into a neutron and a positron is emitted.
$$
\mathrm{p}^{+} \longrightarrow \mathrm{n}^0+\mathrm{e}^{+}
$$
no. of neutrons initially was $A-Z$
no. of neutrons after decay $(A-Z)-3 \times 2$ (due to alpha particles) $+2 \times 1$ (due to positive beta decay)
The no. of proton will reduce by 8 . [as $3 \times 2$ (due to alpha particles) $+2$ (due to positive beta decay)] Hence atomic number reduces by 8 .

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2010
A radioactive nucleus (initial mass number A and atomic number Z) emits three Î± - particles and two positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Options:
A A-Z-4Z-2
B A-Z-8Z-4
C A-Z-4Z-8
D A-Z-12Z-4

The correct answer is:
A-Z-4Z-8

Solution:
After emitting three Î± - particles, the radioactive nucleus (initial mass number A and atomic number Z) will be rendered into a nucleus having mass number (A-12) and atomic number (Z-6). Further, after emitting two positrons, the mass number will remain unchanged i.e., (A-12) but atomic number will become (Z-6-2) i.e., (Z-8).
Therefore, number of protons in the final nucleus =A-12-Z-8=A-Z-4
Therefore, the ratio of number of neutrons to that of protons in the final nucleus
=A-Z-4Z-8

Topic: Physics, Nuclear Physics, NEET
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After $5 \mathrm{~min}-$ utes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is

Options:
A
$0.4 \ln 2$

B
$0.2 \ln 2$

C
$0.1 \ln 2$

D
None of these

The correct answer is:
None of these

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2003
A radioactive sample at some instant has a disintegration rate of 5000 disintegrations per minute. After 5 minutes, the rate becomes 1250 disintegrations per minute. Then, the decay constant (per minute) is

Options:
A 0.4 ln 2
B 0.2 ln 2
C 0.1 ln 2
D 0.8 ln 2

The correct answer is:
0.4 ln 2

Solution:
A A 0 = 1 2 5 0 5 0 0 0 = 1 4 = 1 2 t T
⇒ T = t 2 = 5 2
λ=ln2T=25ln2
=0.4 ln2

Topic: Physics, Atomic Physics, JEE Main
A radioactive sample contains 2.2 mg of pure C611 which has a half-life period of 1224 seconds. Calculate the number of atoms present initially, and the activity when 5 Î¼g of the sample will be left.

Options:
A

2.1× 10 20  atoms,  3.3× 10 14  dps

B

3.1× 10 19  atoms,  2.6× 10 13  dps

C

2.3× 10 19   atoms,  1.6× 10 13  dps

D

1.2Ã 10 20 âatoms,ââ1.55Ã 10 14â dps

The correct answer is:

1.2Ã 10 20 âatoms,ââ1.55Ã 10 14â dps

Solution:
Number of atoms present initially

= 6.023Ã 10 23 Ã2.2Ã 10 â3 11

=1.2Ã 10 20 âatoms

No. of atoms present in 5 Î¼g of the sample

N= 6.023Ã 10 23 Ã5Ã 10 â6 11 =2.74Ã 10 17 atoms

Activity of the sample =Î»âN

= 0.693 T 1 2 ÃN

= 0.693Ã2.74Ã 10 17 1224

=1.55Ã1014âdisintegrations/second

Topic: Physics, Nuclear Physics, JEE Main
A radioactive sample contains two different types of radioactive nuclei A-with half-life 5.0 days and B-with half-life 30.0 days initially the decay rate of the A-type nuclei is 64 time that of B type of nuclei. Their decay rates will be equal when time is 9n days Find the value of n

Options:

The correct answer is:
4

Solution:
λA=ln25 day-1 and λB=ln230 day-1
Given dBOAdt=64dNOSdt
∴ λANOA=64λBNOB
At time t,dNAdt=dNBdt
∴ λANA=λBNB
Dividing equation 1 and 2 NOANA=64 NOBNB
But NA=NOAe-λAt and NB=NOBe-λBt
∴ equation 3becomes eλAt=64eλBt
∴ eλA-λBt=64
∴ λA- λBt= ln64
∴ t=ln64ln25-ln230=36 days

Topic: Physics, Nuclear Physics, JEE Main
A radioactive sample decays 78 times its original quantity in 15 minutes. The half-life of the sample is

Options:
A 5 min
B 7.5 min
C 15 min
D 30 min

The correct answer is:
5 min

Solution:
Remaining amount of sample is=1-78=18

Using radioactive decay law

N=N02-tT12

We have, 2tT12=N0N=N0N08=8=23

Then, tT12=3

Thus, half life of the sample is T12=5 min.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A radioactive sample has an activity \(A\) in air. If the sample is kept inside water, then its activity \(A^{\prime}\)

Options:
A becomes less than \(A\)
B becomes more than \(A\)
C is same as \(A\)
D will be less than or equal to \(A\) depending on the density of water

The correct answer is:
is same as \(A\)

Solution:
Activity of a radioactive sample does not depend on the type of medium in which it is kept. Hence, if the radioactive sample is kept inside water, then its activity \(\left(A^{\prime}\right)\) remains same as in air. i.e.,
\(A^{\prime}=A\)

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 2)
A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 μF is first charged and later connected with resistor R. If the ratio of the charge on the capacitor to the activity of the radioactive sample is fixed with respect to time then the value of R should be_______Ω.

Options:

The correct answer is:
150

Solution:
Tm=30 ms
C=200 μF
qN=Q0e-tRCN0e-λ,t=Q0 N0eλ-1RC

Since qN is constant hence
λ=1RC
R=1λC=TmC=30×10-3200×10-6=150 Ω

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A radioactive sample has half-life of 5 years. The percentage of fraction decayed in 10 years will be

Options:
A $25 \%$
B $50 \%$
C $75 \%$
D $100 \%$

The correct answer is:
$75 \%$

Solution:
Given:
Total time, $\mathrm{T}=10$ years
Half life, $\mathrm{T}_{1 / 2}=5$ years
$\therefore \quad$ No. of half lives, $\mathrm{n}=\frac{10}{5}=2$
From $\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$
$\therefore \quad$ After 10 years, $\frac{1}{4}^{\text {th }}$ of the original substance will remain.
$\Rightarrow\left(1-\frac{1}{4}\right)=\frac{3}{4} \times 100=75 \%$ of the fraction would get decayed.

Topic: Physics, Nuclear Physics, MHT CET, MHT CET 2023 (13 May Shift 1)
A radioactive sample is undergoing α decay, At any time t1, its activity is A and another time t2, the activity is A5. What is the average life time for the sample ?

Options:
A ln5t2-t1
B lnt2+t12
C t2-t1ln5
D t1-t2ln5

The correct answer is:
t2-t1ln5

Solution:
Let initial activity be A0

A=A0e-λt ...1

A5=A0e-λt2 ...2

1÷2

5=eλt2-t1

λ=ln5t2-t1=1τ

τ=t2-t1ln5

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A radioactive sample of half-life 10 days, contains 1000x nuclei. Number of original nuclei present after 5 days is

Options:
A 707x
B 750x
C 500x
D 250x

The correct answer is:
707x

Solution:
The number of radioactive nuclei left after n half lives is
N=N02n
for 5 days n=12
So, N=1000x212=1000x2
=1000x1.414=0.7071000x
=707x

Topic: Physics, Nuclear Physics, JEE Main
A radioactive sample $S_1$ having an activity of $5 \mu \mathrm{Ci}$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu \mathrm{Ci}$. The half lives of $S_1$ and $S_2$ can be

Options:
A
20 yr and 5 yr, respectively

B
20 yr and 10 yr, respectively

C
10 yr each

D
5 yr each

The correct answer is:
20 yr and 5 yr, respectively

Solution:
Activity of $S_1=\frac{1}{2}$ (activity of $S_2$ )
or
$$
\lambda_2 N_1=\frac{1}{2}\left(\lambda_2 N_2\right) \text { or } \frac{\lambda_1}{\lambda_2}=\frac{N_2}{2 N_1}
$$
or
$$
\begin{aligned}
& \frac{T_1}{T_2}=\frac{2 N_1}{N_2}\left(T=\text { half-life }=\frac{\ln 2}{\lambda}\right) \\
& N_1=2 N_2 \\
& \frac{T_1}{T_2}=4
\end{aligned}
$$
Given
$\therefore$ correct option is (a).

Topic: Physics, Nuclear Physics, JEE Main
A radioactive sample S1 having the activity A1 has twice the number of nuclei as another sample S2 of activityA2. If A2=2A1, then the ratio of half-life of S1 to the half-life of S2 is

Options:
A 4
B 2
C 0.25
D 0.75

The correct answer is:
4

Solution:
Activity, A=λN=0.693T1/2 N
Where T1/2 is the half-life of a radioactive sample,
∴ A1A2=N1T1 ×T2N2
T1T2=A2A1 ×N1N2
=2A1A1 ×2N2N2=41

Topic: Physics, Nuclear Physics, JEE Main
A radioactive substance ${ }_{88} X^{228}$ (IIA) emits $3 \alpha$ and $3 \beta$ particles to form " $Y$ ". To which group of long form of the periodic table does " $Y$ " belong ?

Options:
A IVA
B VA
C VIA
D VIIA

The correct answer is:
VIIA

Solution:
${ }_{88}^{88} X^{228} \longrightarrow{ }_{\text {IIA group }}^{85} Y^{216}+3 \alpha+3 \beta$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
A radioactive substance decays to 116th of its initial activity in 80 days. The half life of the radioactive substance expressed in days is ________.

Options:

The correct answer is:
20

Solution:
N0⟶t12N02⟶t12N04⟶t12N08⟶t12N016

4×t12=80

t12=20 days

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A radioactive substance of half life 138.6 days is placed in a box. After \(n\) days only \(20 \%\) of the substance is present then the value of \(n\) is \([\ln (5)=1.61]\)

Options:
A 693
B 138.6
C 277.2
D 322

The correct answer is:
322

Solution:
Half life of a radioactive substance, \(n_{1 / 2}=138.6\) days
If \(N_0\) be the initial amount of radioactive substance, then remaining amount after \(n\) days is given by
\(N=20 \% \text { of } N_0=\frac{20}{100} \times N_0=\frac{N_0}{5}\)
By radioative decay's law,
\(N=N_0\left(\frac{1}{2}\right)^{\frac{n}{n_{1 / 2}}} \Rightarrow \frac{N_0}{5}=N_0\left(\frac{1}{2}\right)^{\frac{n}{138.6}} \Rightarrow \frac{1}{5}=\left(\frac{1}{2}\right)^{\frac{n}{138.6}}\)
Taking \(\log\) on the both sides, we get
\(\begin{aligned}
\ln \frac{1}{5} & =\ln \left(\frac{1}{2}\right)^{\frac{n}{138.6}} \Rightarrow \ln \frac{1}{5}=\frac{n}{138.6} \ln \left(\frac{1}{2}\right) \\
\ln 5 & =\frac{n}{138.6} \ln 2 \\
n & =138.6 \times \frac{\ln 5}{\ln 2} \quad \left[\begin{array}{l}\because \ln 5=1.61 \\ \ln 2=0.693\end{array}\right]\\
n & =138.6 \times \frac{1.61}{0.693} \Rightarrow n=322 \text { days }
\end{aligned}\)

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)