datavorous/entrance-exam-dataset huggingface dataset:
The $\mathrm{H}-\mathrm{C}-\mathrm{H}, \mathrm{H}-\mathrm{N}-\mathrm{H}$, and $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angles (in degrees) in methane, ammonia and water are respectively, closest to

Options:
A $109.5,104.5,107.1$
B $109.5,107.1,104.5$
C $104.5,107.1,104.5$
D $107.1,104.5,109.5$

The correct answer is:
$109.5,107.1,104.5$

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The $\mathrm{H}-\mathrm{H}$ bond energy is $430 \mathrm{~kJ} \mathrm{~mol}$ and $\mathrm{Cl}-\mathrm{Cl}$ bond energy is $240 \mathrm{~kJ} \mathrm{~mol}^{-1} . \Delta \mathrm{H}$ for $\mathrm{HCl}$ is $-90 \mathrm{~kJ}$.
The $\mathrm{H}-\mathrm{Cl}$ bond energy is about

Options:
A $180 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $360 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $213 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $425 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$425 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \rightarrow \mathrm{HCl}, \Delta \mathrm{H}=-90 \mathrm{KJ}$
$\therefore \Delta H=\frac{1}{2} E_{H-H}+\frac{1}{2} E_{C-C]}$
or $-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-E_{H C}$
$\therefore E_{H-C}=425 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
The $\mathrm{H}-\mathrm{O}-\mathrm{O}$ bond angle in $\mathrm{H}_2 \mathrm{O}_2$ is

Options:
A $107.28^{\circ}$
B $109.28^{\circ}$
C $104.5^{\circ}$
D $97^{\circ}$

The correct answer is:
$97^{\circ}$

Solution:

Topic: Chemistry, Hydrogen, JEE Main
The haemoglobin and the gold sol are examples of

Options:
A positively charged sols,
B negatively and positively charged sols, respectively.
C negatively charged sols.
D positively and negatively charged sols, respectively.

The correct answer is:
positively and negatively charged sols, respectively.

Solution:
Hemoglobin and gold sol both are colloids and always carry an electric charge. Hemoglobin is a positively charged sol, because in hemoglobin, Fe+2 ion is the central metal ion of the octahedral complex. All metal sols like, Au-sol, Ag-sol etc. are negatively charged sols

The gold sol is negatively charged, while the haemoglobin is positively charged.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The haemoglobin content per $100 \mathrm{~mL}$ of blood of a normal healthy human adult is

Options:
A 5-11 $g$
B $25-30 \mathrm{~g}$
C $17-20 \mathrm{~g}$
D $12-16 \mathrm{~g}$

The correct answer is:
$12-16 \mathrm{~g}$

Solution:
The normal haemoglobin count is about 15 $\mathrm{gm} / 100 \mathrm{~mL}$ of blood in males, while for females the haemoglobin count is slightly less, ie, $13 \mathrm{gm} / 100 \mathrm{~mL}$ blood.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2010 (Mains)
The haemoglobin from red blood corpuscles of most mammals contains approximately 0.33% of iron by mass. The molecular mass of haemoglobin is 67200. The number of iron atoms in each molecular of heamoglobin is [Given : Atomic mass of Fe = 56 g/mol]
Report your answer by rounding it upto nearest whole number.

Options:

The correct answer is:
4

Solution:
100 g Haemoglobin =0.33 g Fe
67200 g Haemoglobin =0.33×67200100=221.76
Number of iron atoms per molecules of haemoglobin =221.7656=4

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The haemoglobin of a human foetus

Options:
A Has a higher affinity for oxygen than that of an adult
B Has a lower affinity for oxygen than that of the adult
C Its affinity for oxygen is the same as that of an adult
D Has only 2 protein subunits instead of 4

The correct answer is:
Has a higher affinity for oxygen than that of an adult

Solution:
The haemoglobin of human foetus $(\mathrm{HbF})$ has 4 Heme group and 1 globin. In $\mathrm{HbF}$ the globin protein is made up of $2^\alpha$ and $2 \beta$ polypeptide chains. Each $\alpha$ has 141 amino acids and each $\beta$-has 146 amino acids but 37 amino acids in each $\beta$ are different from that of $\mathrm{HbA}$. It has higher affinity for $\mathrm{O}_2$ even at low $\mathrm{PO}_2$ and the oxygen $-\mathrm{Hb}$ dissociation curve remains towards the left of normal.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2008 (Mains)
The half life of a first order reaction is $69.35 \mathrm{sec}$. The value of the rate constant of the reaction is

Options:
A $1.0 \mathrm{~s}^{-1}$
B $0.1 \mathrm{~s}^{-1}$
C $0.01 \mathrm{~s}^{-1}$
D $0.001 \mathrm{~s}^{-1}$

The correct answer is:
$0.01 \mathrm{~s}^{-1}$

Solution:
$k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{69.35}=9.99 \times 10^{-3}=0.01 \mathrm{~s}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The half life of a radioactive nucleus is 50 days. The time interval $\left(t_2-t_1\right)$ between the time $t_2$ when $\frac{2}{3}$ of it has decayed and the time $t_1$ when $\frac{1}{3}$ of it had decayed is

Options:
A 30 days
B 50 days
C 60 days
D 15 days

The correct answer is:
50 days

Solution:
Active fraction at instant $t_2$,
$$
\frac{1}{2^{t_2 / T_{1 / 2}}}=\frac{1}{3}
$$
Active fraction at instant $t_1$,
$$
\begin{aligned}
\frac{1}{2^{t_1 / T_{1 / 2}}} & =\frac{2}{3} \\
\frac{2^{t_2 / T_{1 / 2}}}{2^{t_1 / T_{1 / 2}}} & =2 \\
t_2-t_1=T_{1 / 2} & =50 \text { days }
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2012 (Mains)
The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6.25% of its initial value. The value of x is ____

Options:

The correct answer is:
20

Solution:
6.25% of the initial value means, final number of nuclei should be, N016.

Number of nuclei remaining after n half lives is N=N02n
6.25% N0=N02n
6.25100N0=N02n ⇒625104=12n
⇒2n=10454 ⇒2n=24
⇒n=4

Therefore, time taken will be, n×t12=4×5=20 years

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The half life of a radioactive substance is 20 minutes. The approximate time interval $\left(t_2-t_1\right)$ between the time $t_2$ when $\frac{2}{3}$ of it has decayed and time $t_1$ and $\frac{1}{3}$ of it had decayed is :

Options:
A
$14 \mathrm{~min}$

B
$20 \mathrm{~min}$

C
$28 \mathrm{~min}$

D
$7 \mathrm{~min}$

The correct answer is:
$20 \mathrm{~min}$

Solution:
$\mathrm{t}_{\frac{1}{2}}=20$ minutes
$\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_2} \quad \lambda \mathrm{t}_1=\ln 3$
$\frac{2}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_2} \mathrm{t}_1=\frac{1}{\lambda} \ln 3$
$\frac{2}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda t_2}$
$\mathrm{t}_2=\frac{1}{\lambda} \ln \frac{3}{2}$
$\mathrm{t}_2-\mathrm{t}_1=\frac{1}{\lambda}\left[\ln \frac{3}{2}-\ln 3\right]$
$=\frac{1}{\lambda} \ln \left[\frac{1}{2}\right]=\frac{0.693}{\lambda}$
$=20 \min$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2011
The half life of a radioactive substance is 3.6 days. How much of 20 mg of this radioactive substance will remain after 36 days?

Options:
A 0.0019 mg
B 1.019 mg
C 1.109 mg
D 0.019 mg

The correct answer is:
0.019 mg

Solution:
Half life $T_{1 / 2}=3.6$ days
Initial quantity $N_0=20 \mathrm{mg}$
Total time $=36$ days
The number of half lives
$n=\frac{t}{T_{1 / 2}}=\frac{36}{3.6}=10$
Hence, mass of radioactive substance left after 10 half lives
$N=N_0 \times\left(\frac{1}{2}\right)^n=20 \times \frac{1}{1024}=0.019 \mathrm{mg}$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2004
The half life of a radioactive substance is T. The time taken, for disintegrating 78th part of its original mass will be:

Options:
A 2T
B 3T
C T
D 8T

The correct answer is:
3T

Solution:
It is given that t12= T.

After disintegration of 78thpart, 18th part is only left.

Using the decay formula,

N=N012tt12

⇒N08=N012tT⇒-ln8=-tTln2⇒3ln2=tTln2⇒t=3T

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The half - life of a radioisotope is four hours. If the initial mass of the isotope was $200 \mathrm{~g}$, the mass remaining after 24 hours undecayed is

Options:
A
$1.042 \mathrm{~g}$

B
$4.167 \mathrm{~g}$

C
$3.125 \mathrm{~g}$

D
$2.084 \mathrm{~g}$

The correct answer is:
$3.125 \mathrm{~g}$

Solution:
$2^6=\frac{200}{a-x}$
$(a-x)=3.125 \mathrm{gm}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2004
The half life of a radium is about 1600 years. Of $100 \mathrm{~g}$ of radium existing now 25 g will remain unchanged after :-

Options:
A 4800 years
B 6400 years
C 2400 years
D 3200 years

The correct answer is:
3200 years

Solution:
Applying the formula
$$
\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow \quad \frac{N}{N_0} & =\left(\frac{1}{2}\right)^n
\end{aligned}
$$
$$
\Rightarrow \frac{25}{100}=\left(\frac{1}{2}\right)^n \Rightarrow n=2
$$
$\therefore$ Total time is $2 \times 1600=3200$ year

Topic: Physics, Nuclear Physics, NEET, NEET 2004
The half life of a stream of radioactive particles moving along a straight path with a constant kinetic energy of $4 \mathrm{eV}$ is 1 minute. The percentage of particles which decay before travelling a distance of $3.6 \mathrm{~km}$ is (Mass of the radioactive particles $=3.2 \times 10^{-21} \mathrm{~kg}$ and charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$ ).

Options:
A 87.5
B 175
C 37.5
D 75

The correct answer is:
87.5

Solution:
$\begin{aligned}
& \text {As, } K . E .=\frac{1}{2} m v^2 \Rightarrow \mathrm{V}=\sqrt{\frac{2 \times K . E .}{m}} \\
& \Rightarrow \mathrm{V}=\sqrt{\frac{2 \times 4 \times 16 \times 10^{-19}}{32 \times 10^{-21}}}=\sqrt{\frac{2 \times 4}{2 \times 10^{-2}}} \\
& \Rightarrow \mathrm{V}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Time to travel a distance of $3.6 \mathrm{~km}$ is
$\begin{aligned}
& t=\frac{D}{v}=\frac{3.6 \times 1000}{20} s=\frac{3.6 \times 1000}{20 \times 60} \mathrm{~min} \\
& =3 \mathrm{~min}=3 \text { half-time }
\end{aligned}$
In three half life times, the number of particles decayed
$\begin{aligned}
& =\mathrm{N}_0-\frac{\mathrm{N}_0}{2^3}=\frac{7}{8} \mathrm{~N}_0 \\
& \% \text { req. }=\frac{\frac{7}{8}}{\mathrm{~N}_0} \mathrm{~N}_{\mathrm{o}} \times 100=87.5 \%
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
The half life of neutron is 693 seconds. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of 0.084 $\mathrm{eV}$, travells a distance of $1 \mathrm{~km}$ ?
(mass of neutron $=1.68 \times 10^{-27} \mathrm{~kg}$, and $\ln 2=0.693$ )

Options:
A $0.25 \times 10^{\wedge}-5$
B $0.5 \times 10^{\wedge}-5$
C $0.8 \times 10^{\wedge}-5$
D $10^{\wedge}-5$

The correct answer is:
$0.25 \times 10^{\wedge}-5$

Solution:
Given, the half-life of neutron, $t_{1 / 2}=693 \mathrm{sec}$ Kinetic energy of neutron $=0.084 \mathrm{eV}$
$\begin{aligned}
\frac{1}{2} m v^2 & =0.084 \times 1.6 \times 10^{-19} \\
v^2 & =\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{m} \\
& =\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{1.68 \times 10^{-27}}=0.16 \times 10^8 \\
v & =0.4 \times 10^4 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Time taken by neutron to travelled in $1 \mathrm{~km}$,
$t=\frac{1000}{0.4 \times 10^4} \Rightarrow t=0.25 \mathrm{~s}$
By Radioactive Decay's law,
$\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N}{N_0}=\left(\frac{1}{2}\right)^n \\
\frac{N_0}{N} & =2^n
\end{aligned}$
Taking log on the both sides, we get
$\begin{aligned}
\ln \frac{N_0}{N}=\ln 2^n & =n \ln 2=\frac{t}{t_{1 / 2}} \ln 2=\frac{0.25 \times 0.693}{693} \\
& =0.25 \times 10^{-5} \quad[\because \ln 2=0.693]
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
The half life of radioactive Radon is $3.8$ days.
The time at the end of which $\frac{1}{20}$ th of the radon sample will remain undecayed is (given $\left.\log _{10} e=0.4343\right)$

Options:
A $3.8$ days
B $16.5$ days
C 33 days
D 76 days.

The correct answer is:
$16.5$ days

Solution:
$\begin{array}{l}
t_{1 / 2}=3.8 \text { day } \\
\therefore \quad \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{3.8}=0.182
\end{array}$
If the initial number of atom is $a=A_{0}$ then after time $t$ the number of atoms is $a / 20=A$. We have to find $t$.
$$
\begin{aligned}
t &=\frac{2.303}{\lambda} \log \frac{A_{0}}{A}=\frac{2.303}{0.182} \log \frac{a}{a / 20} \\
&=\frac{2.303}{0.182} \log 20=16.46 \text { days }
\end{aligned}
$$

Topic: Physics, Nuclear Physics, VITEEE, VITEEE 2016
The half life of tritium is \( 12.5 \) years. What mass of tritium of initial mass \( 64 \mathrm{~g} \) will remain
undecayed after \( 50 \) years?

Options:
A \( 32 \mathrm{mg} \)
B \( 8 \mathrm{mg} \)
C \( 16 \mathrm{mg} \)
D \( 4 \mathrm{mg} \)

The correct answer is:
\( 4 \mathrm{mg} \)

Solution:
Given, half-life of tritium $=12.5$ years; initial mass $=64 \mathrm{mg} ;$ time, $\mathrm{t}=50$ years
We know that
$N=N_{0} e^{-\lambda t}$
and $\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{12.5 \text { years }}$
$\therefore \frac{N}{N_{0}}=e^{\frac{-0.693}{12.5} \times 50}$
$\Rightarrow \frac{N}{N_{0}}=e^{-2.772}=0.0625$
$\Rightarrow N=N_{0} \times 0.0625=64 \times 0.0625 \mathrm{mg}$
$\Rightarrow N=4 \mathrm{mg}$
Thus, 4 mg of tritium will remain undecayed after 50 years

Topic: Physics, Nuclear Physics, KCET, KCET 2018
The half life period of a first order chemical reaction is $6.93$ minutes. The time required for the completion of $99 \%$ of the chemical reaction will be $(\log 2=0.301)$ :

Options:
A
$230.3$ minutes

B
$23.03$ minutes

C
$46.06$ minutes

D
$460.6$ minutes

The correct answer is:
$46.06$ minutes

Solution:
$$
\begin{aligned}
& \because \lambda=\frac{0.6932}{t_{1 / 2}}=\frac{0.6932}{6.93} \min ^{-1} \\
& \text { Also } t=\frac{2.303}{\lambda} \log \frac{\left[A_{\circ}\right]}{[A]} \\
& {\left[A_{\circ}\right]=\text { initial concentration (amount) }} \\
& {[A]=\text { final concentration (amount) }} \\
& \therefore t=\frac{2.303 \times 6.93}{0.6932} \log \frac{100}{1} \\
& =46.06 \text { minutes }
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2009
The half life period of a first order reaction is 1 $\min 40$ secs. Calculate its rate constant.

Options:
A $6.93 \times 10^{-3} \mathrm{~min}^{-1}$
B $6.93 \times 10^{-3} \mathrm{sec}^{-1}$
C $6.93 \times 10^{-3} \mathrm{sec}$
D $6.93 \times 10^{3} \mathrm{sec}$

The correct answer is:
$6.93 \times 10^{-3} \mathrm{sec}^{-1}$

Solution:
For $1^{\text {st }}$ order reaction.
$\mathrm{t}_{1 / 2} \text { (halflife time) }=\frac{0.693}{\mathrm{~K}}$
Hence $\mathrm{K}=\frac{0.693}{\mathrm{t}^{1 / 2}}=\frac{0.693}{(60+40) \mathrm{sec}}=0.693 \times$ $10^{-2} \mathrm{sec}^{-1}=6.93 \times 10^{-3} \mathrm{sec}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The half life period of a first order reaction

Options:
A $\frac{0.693}{t}$
B $\frac{0.693}{K}$
C $\frac{2.303}{t}$
D $\frac{0.303}{K_1}$

The correct answer is:
$\frac{0.693}{K}$

Solution:
Derive ${ }^{t_{1 / 2}}$ from $K_t=2.303 \log \frac{a}{a-x}$, putting ${ }^{t=\frac{1}{2}}$ and $x=\frac{a}{2}$. Therefore it is $\frac{0.693}{K}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The half life period of a radioactive substance is 60 days. The time taken for 78 th of its original mass to disintegrate will be :

Options:
A 120 days
B 130 days
C 180 days
D 20 days

The correct answer is:
180 days

Solution:
Given half life period of a radioactive substance is t12=60 days.

Here, 78th of radioactive substance disintegrates means 18th remains or 123.

∴ 3 half lives =3×60=180 days

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The half life period of a radioactive substance is 140 days. After how much time, 15 g will decay from a 16 g sample of the substance?

Options:
A 140 days
B 560 days
C 420 days
D 280 days

The correct answer is:
560 days

Solution:
Ratio of mass remaining undecayed to initial mass is,
mm0=12tT
16-1516=12tT
124=12tT
⇒ t=4T=4×140 days
t=560 days

Topic: Physics, Nuclear Physics, NEET
The half of a radioactive substance is 25 minutes The time between $50 \%$ decay and $87.5 \%$ decay of the substance will be

Options:
A 75 min
B $25 \mathrm{~min}$
C $37.5 \mathrm{~min}$
D $50 \mathrm{~min}$

The correct answer is:
$50 \mathrm{~min}$

Solution:
For $50 \%$ decay, time is equal to one half-life.
For $87.5 \%$ decay, time is three half - lives.
$\therefore$ Time interval $\Delta \mathrm{t}=2$ half - lives.
$\Delta t=2 \times 25 \mathrm{~min}=50 \mathrm{~min}$

Topic: Physics, Nuclear Physics, JEE Main
The half-life for decay of ${ }^{14} \mathrm{~C, by~} \beta$ -emission is $5730 \mathrm{yr}$. The fraction of ${ }^{14} \mathrm{C}$ decays, in a sample that is 22920 yr old, would be

Options:
A $\frac{1}{8}$
B $\frac{1}{16}$
C $\frac{7}{8}$
D $\frac{15}{16}$

The correct answer is:
$\frac{15}{16}$

Solution:
$n=\frac{\text { total time }(t)}{\text { half -lite }(t / 2)}=\frac{22920}{5730}=4$
$$
\begin{aligned}
\text { Left amount, } N=& N_{0}\left(\frac{1}{2}\right)^{n} \\
&=N_{0}\left(\frac{1}{2}\right)^{4}=\frac{N_{0}}{16} \\
\therefore \text { Decayed fraction } &=N_{0}-\frac{N_{0}}{16} \\
&=\frac{16 N_{0}-N_{0}}{16}=\frac{15 N_{0}}{16}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
The half-life for decay of radioactive 14 C is 5730 years. An archaeological artefact containing wood has only 80% of the 14 C activity as found in living trees. The age of the artefact would be:
[Given: log 1.25 = 0.0969]

Options:
A 1845 years
B 184.5 years
C 1900 years
D 190 years

The correct answer is:
1845 years

Solution:
Radioactive decay follows first order kinetics. Letâ [A] 0 =100

â´â[A]=100Ã80%=80 Decayâconstantââ(k)= 0.693 t 1/2 = 0.693 5730 t= 2.303 k âlog [Ao] [A] = 2.303 0.693 5730 âlog 100 80 = 2.303Ã5730 0.693 Ãâlog1.25 = 2.303Ã5730 0.693 Ã0.0969 = 1 8 4 5 yrs

Topic: Chemistry, Chemical Kinetics, JEE Main
The half-life of a radioactive element $\mathrm{A}$ is the same as the mean-life of another radioactive element B. Initially both substances have the same number of atoms, then :

Options:
A
A and B decay at the same rate always.

B
A and $\mathrm{B}$ decay at the same rate initially.

C
A will decay at a faster rate than B.

D
$\mathrm{B}$ will decay at a faster rate than $\mathrm{A}$.

The correct answer is:
$\mathrm{B}$ will decay at a faster rate than $\mathrm{A}$.

Solution:
$\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}=\left(\mathrm{t}_{\text {mean }}\right)_{\mathrm{B}}$
$$
\Rightarrow \frac{0.693}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=0.693 \lambda_{\mathrm{B}}
$$
or $\lambda_{\mathrm{A}} < \lambda_{\mathrm{B}}$
Also rate of decay $=\lambda \mathrm{N}$
Initially number of atoms $(\mathrm{N})$ of both are equal but since $\lambda_{\mathrm{B}}>\lambda_{\mathrm{A}}$, therefore $\mathrm{B}$ will decay at a faster rate than $\mathrm{A}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2013 (22 Apr Online)
The half-life of a radioactive element is $10 \mathrm{~h}$. The fraction of initial radioactivity of the element that will remain after $40 \mathrm{~h}$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{16}$
C $\frac{1}{8}$
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{16}$

Solution:
After 4 half-lifes
$\frac{N_0}{2^4}=\frac{N_0}{16}$

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2012
The half-life of a radioactive element is $10 \mathrm{~h}$. The fraction of initial radioactivity of the element that will remain after $40 \mathrm{~h}$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{16}$
C $\frac{1}{8}$
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{16}$

Solution:
After 4 half-lifes
$\frac{N_0}{2^4}=\frac{N_0}{16}$

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2012
The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of it in grams remaining undecayed after 18 hours would be:

Options:

The correct answer is:
4

Solution:
t1/2=3 hours, n=T/t1/2=18/3=6
N = N 0 1 2 6
= 2 5 6 Ã 1 8 Ã 1 8 = 4 g Â·

Topic: Chemistry, Chemical Kinetics, JEE Main
The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were $256 \mathrm{~g}$, the mass of it remaining undecayed after 18 hours would be

Options:
A
$8.0 \mathrm{~g}$

B
$12.0 \mathrm{~g}$

C
$16.0 \mathrm{~g}$

D
$4.0 \mathrm{~g}$

The correct answer is:
$4.0 \mathrm{~g}$

Solution:
$\mathrm{t}_{1 / 2}=3 \mathrm{hrs}$. Initial mass $\left(\mathrm{C}_0\right)=256 \mathrm{~g}$
$\therefore \mathrm{C}_{\mathrm{n}}=\frac{\mathrm{C}_0}{2^{\mathrm{n}}}=\frac{256}{(2)^6}=\frac{256}{64}=4 \mathrm{~g}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2003
The half-life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was estimated to be

Options:
A 150 years
B 200 years
C 250 years
D 100 years

The correct answer is:
200 years

Solution:
N⁡ N⁡ 0 = 1 2 n
where n is number of half-lives
∴116=12nor124=12n orn=4
Let the age of rock be t years
∵ n = t T⁡ 1 / 2
or t= nT1/2 = 4 × 50 years = 200 years

Topic: Physics, Nuclear Physics, JEE Main
The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is :

Options:
A 18
B 14
C 78
D 34

The correct answer is:
78

Solution:
The formula to calculate the number of half lives covered by the radioactive material is given by

n=tT12 ...1

Substitute the values of the known parameters into equation (1) to calculate the number of half lives covered.

n=15 years5 years= 3

The relation between the amount of radioactive substance after certain time can be written as

N=N02-n ...2

Substitute the value of n into equation (2) to obtain the fraction of the remaining substance.

N=N02-3⇒NN0=18

Hence, the amount of the decayed substance can be written as

1-NN0=1-18=78

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The half-life of a radioactive nuclide is 20 h. It is found that the fraction (1x) of original activity remains after 40 hours? What is the value of x?

Options:

The correct answer is:
4

Solution:
We have t t 1 / 2 = 4 0 hours 20 hours = 2
Thus, A = A 0 2 t / t 1 / 2 = A 0 2 2 = A 0 4
So, one fourth of the original activity will remain after 40 hours.

Topic: Physics, Nuclear Physics, JEE Main
The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :

Options:
A 12
B 122
C 23
D 232

The correct answer is:
122

Solution:
Given,

T12=100 hours

t=150 hours

As we know that the number of nuclide remain after n half life is given by,

N=N02n

∵100 hours =1 half life

150 hours =1100×150

=32 half life

So, Fraction remain after 32 half life

NN0=123/2=122

Topic: Physics, Nuclear Physics, NEET, NEET 2021
The half-life of a radioactive sample is \(T\). The fraction of the initial mass of the sample that decays in an interval \(T / 2\) is

Options:
A \(\frac{1}{\sqrt{2}}\)
B \(\sqrt{2}\)
C \(\frac{(\sqrt{2}-1)}{\sqrt{2}}\)
D \(\frac{(\sqrt{2}+1)}{\sqrt{2}}\)

The correct answer is:
\(\frac{1}{\sqrt{2}}\)

Solution:
Fraction remains after \(n\) half-lives is given as
\(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{t / T} \quad\) [where, \(T\) is half-life]
\(\Rightarrow \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}\)
Given, \(\quad t=\frac{T}{2}\)
\(\therefore \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{T / 2}{T}}=\left(\frac{1}{2}\right)^{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)

Topic: Physics, Nuclear Physics, JEE Main
The half-life of a radioactive sample is \(T\). The fraction of the initial mass of the sample that decays in an interval \(T / 2\) is

Options:
A \(\frac{1}{\sqrt{2}}\)
B \(\sqrt{2}\)
C \(\frac{(\sqrt{2}-1)}{\sqrt{2}}\)
D \(\frac{(\sqrt{2}+1)}{\sqrt{2}}\)

The correct answer is:
\(\frac{1}{\sqrt{2}}\)

Solution:
Fraction remains after \(n\) half-lives is given as
\(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{t / T} \quad\) [where, \(T\) is half-life]
\(\Rightarrow \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{t}{T}}\)
Given, \(\quad t=\frac{T}{2}\)
\(\therefore \quad \frac{N}{N_0}=\left(\frac{1}{2}\right)^{\frac{T / 2}{T}}=\left(\frac{1}{2}\right)^{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)