datavorous/entrance-exam-dataset huggingface dataset:
The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

Options:
A

B

C

D

The correct answer is:

Solution:
de-Broglie wavelength $\lambda=\frac{h}{P}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}$ where $E=\frac{1}{2} m v^2$
Squaring both sides,
$\begin{aligned}
& \lambda^2=\frac{h^2}{4 m^2 E} \\
& \Rightarrow \frac{1}{\lambda^2}=\text { (constant) } E
\end{aligned}$
Graph passes through origin with constant slope.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2024
The graph which shows the variation of the de Broglie wavelength λ of a particle and its associated momentum p is :

Options:
A
B
C
D

The correct answer is:

Solution:
As per de Broglie,

λ=hp which is similar to y∝1xexpression.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2022 (Phase 1)
The gravitational field in a region is given by gâ=5i^+12j^ N kg-1. The change in the gravitational potential energy of a particle of mass 2 kg when it is taken from the origin to a point 7 m,-3 m is

Options:
A 71ââJ
B 1 3 5 8 J
C 2ââJ
D 1  J

The correct answer is:
2  J

Solution:
dV=-Eâ.drâ

ââV=-Eâ.ârâ
=-5i^+12j^Â·7i^-3j^
= - 3 5 - 3 6 = 1
âÎU=mÎV=2J

Topic: Physics, Gravitation, NEET
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig. )

Options:
A a,
B b,
C c,
D 0.

The correct answer is:
c,

Solution:
Correct Option is : (C)
c,

Topic: Chemistry, Gravitation, NEET
The Great Barrier Reef along the east coast of Australia can be categorised as

Options:
A Population
B Community
C Ecosystem
D Biome

The correct answer is:
Ecosystem

Solution:
The great Barrier Reef - the largest system of coral reefs in the world - is more than 2,000 km in length and comprises 2,900 separate reefs and 940 islands. Its high species diversity includes more than 400 species of coral, 4,000 species of molluscs, 1,500 species of fish, 6 species of turtles, 35 species of seabirds and 23 species of sea mammals, The Great Barrier Reef is one of the richest and most complex natural systems in the world. The Reef's diverse range of habitats all connect and support each other, which means they are all vital to the life of the Great barrier Reef

Topic: Biology, Ecosystem, NEET
The 'greenhouse effect' results in

Options:
A the warming of climate
B blocking Earth's heat from escaping into space
C increases the temperature of Earth, which sustains life
D all of the above

The correct answer is:
all of the above

Solution:
The 'greenhouse effect' is a climate warming effect that results when the atmosphere traps heat radiating from Earth towards space. When solar radiation falls on the earth, it re-emits heat in the form of infra-red radiation. However, part of this reflected heat does not leave the Earth's atmosphere and is trapped by gases such as carbon dioxide, methane, etc. This increases the temperature of the Earth. Without these gases, the temperature of the Earth would be -18°C.

Topic: Biology, Environmental Issues, NEET
The gritty nature of diatomaceous earth is due to:

Options:
A Silica in the cell wall
B Ability to photosynthesize
C Calcium in the cell wall
D Excess salt in the cell from the sea

The correct answer is:
Silica in the cell wall

Solution:
In diatoms, the cell walls form two thin overlapping shells, which fit together as in a soapbox. The walls are embedded with silica and thus, the walls are indestructible. Hence, diatoms have left behind a large amount of cell wall deposits in their habitat; this accumulation over billions of years is referred to as ‘diatomaceous earth’. Being gritty, this soil is used in polishing, filtration of oils and syrups.

Topic: Biology, Biological Classification, NEET
The ground state energy of H-atom is 13.6 eV. The energy needed to ionize H-atom from its second excited state is

Options:
A 1.51 eV
B 3.4 eV
C 13.6 eV
D 12.1 eV

The correct answer is:
1.51 eV

Solution:
Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

for this case, initially, the electron is in 3rd orbit and we know for H-atom En=3=-1.51 eV
Hence ionization energy of the electron is 1.51 eV

Topic: Physics, Atomic Physics, NEET
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. The energy needed to ionize hydrogen atom from its second excited state will be

Options:
A $1.51 \mathrm{eV}$
B $3.4 \mathrm{eV}$
C $13.6 \mathrm{eV}$
D $6.8 \mathrm{eV}$

The correct answer is:
$1.51 \mathrm{eV}$

Solution:
We know that total energy of electron in hydrogen atom is given by $E_\eta=\frac{-13.6}{n^2}$
Also, total energy of electron in second excited state $=\frac{-13.6}{(3)^2}=-1.51 \mathrm{eV}$
$\Rightarrow$ lonization energy or energy required to ionize hydrogen atom from its second excited state $=1.51 \mathrm{eV}$

Topic: Physics, Atomic Physics, NEET, NEET 2023 (Manipur)
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. When its electron is in the first excited state, its excitation energy is

Options:
A $3.4 \mathrm{eV}$
B $6.8 \mathrm{eV}$
C $10.2 \mathrm{eV}$
D zero

The correct answer is:
$10.2 \mathrm{eV}$

Solution:
Key Idea : Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (ie, $n=2,3,4 \ldots)$.
Given, ground state energy of hydrogen atom
$$
E_1=-13.6 \mathrm{eV}
$$
Energy of electron in first excited state (ie, $n=2$ )
$$
E_2=-\frac{13.6}{(2)^2} \mathrm{eV}
$$
Therefore, excitation energy
$$
\begin{aligned}
\Delta E & =E_2-E_1 \\
& =-\frac{13.6}{4}-(-13.6) \\
& =-3.4+13.6=10.2 \mathrm{eV}
\end{aligned}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2008 (Screening)
The group of cell initials for tissue culturing is called

Options:
A Callus
B Hybrid
C Sub-culture
D Explant

The correct answer is:
Explant

Solution:
Explants are prepared by dissecting and mincing the tissue into small masses, followed by plating into appropriate tissue culture wells and incubating for a finite period of time.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
The growth of the given system per unit time which is expressed on a common basis or per unit initial parameter is known as

Options:
A Absolute growth rate
B Relative growth rate
C Both a and b
D Exponential growth rate

The correct answer is:
Relative growth rate

Solution:
Quantitative comparisons between the growth of living system can also be made in two ways. Measurement and the comparison of total growth per unit time is called the absolute growth rate and the growth of the given system per unit time expressed on a common basis, per unit initial parameters is called the relative growth rate.

Topic: Biology, Plant - Growth and Development, NEET
The guts of cow and buffalo possess:

Options:
A Fucus sp.
B Chlorella sp.
C Methanogens
D Cyanobacteria

The correct answer is:
Methanogens

Solution:
The Bovine rumen is a niche that has a narrow pH range and is kept fairly stabilized, so that the bacteria in the rumen as well as the intake sacs for food and water are not too disturbed. The organisms responsible for producing methane through bovine rumen are termed methanogens and do so in order to reduce the amount of carbon in the rumen system for fermentation. There are two major divisions within the methanogens found in the bovine rumen. The Methanobrevibacter ruminatium and the Methanosphaera stadtmanae.

Topic: Biology, Biological Classification, NEET, NEET 2015 (Phase 1)
The gynoecium consists of many free pistils in flowers of

Options:
A Aloe
B tomato
C Papaver
D Michelia

The correct answer is:
Michelia

Solution:
Apocarpous condition arises when the number of carpels is two or more and they are free from each other. e.g., Clematis, Michelia (Magnoliaceae), Aconitum, Ranunculus (Buttercup), etc.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2012 (Screening)
The gynoecium may have more than one pistil. Such a condition is called

Options:
A Monocarpellary
B Apocarpous
C Syncarpous
D Multicarpellary

The correct answer is:
Multicarpellary

Solution:
The gynoecium represents the female reproductive part of the flower. The gynoecium may consist of a single pistil (monocarpellary) or may have more than one pistil (multicarpellary).

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The [H+] of a resulting solution that is 0.01 M acetic acid (Ka = 1.8 x 10-5) and 0.01 M in benzoic acid (Ka = 6.3 x 10-5) is

Options:
A 9 x 10-4
B 81 x 10-4
C 9 x 10-5
D 2.8 x 10-3

The correct answer is:
9 x 10-4

Solution:
H + = k 1 C 1 + k 2 C 2
(for mixture of two weak acids)
= 1.8 Ã 1 0 - 5 Ã 0.01 + 6.3 Ã 1 0 - 5 Ã 0.01
= 18 Ã 1 0 - 8 + 6 3 Ã 1 0 - 8
= 9 Ã 1 0 - 4

Topic: Chemistry, Ionic Equilibrium, NEET
The habit of a cabbage or acaulescent plant can be changed drastically by the application of

Options:
A IAA
B GA3
C ABA
D Zeatin

The correct answer is:
GA3

Solution:
Gibberellins promote bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Topic: Biology, Plant - Growth and Development, NEET
The haemoglobin of a human foetus

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

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

Solution:
Haemoglobin is the protein that makes red blood corpuscles red and binds easily and reversibly with oxygen. Normal values for haemoglobin are $14-20 \mathrm{~g} / 100 \mathrm{~mL}$ of blood in infants, $13-18 \mathrm{~g} / 100 \mathrm{~mL}$ in adult male and $12-16 \mathrm{~g} / 100 \mathrm{~mL}$ in adult females. Haemoglobin is made up of protein globin bound to the red haem pigment. The globin consists of four polypeptide chains-2 alpha and 2 beta. Foetal haemoglobin does not sickle even in those distined to have sickle cell anaemia, $i e$, haemoglobin of foetus has higher affinity of oxygen than that an adult.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2008 (Screening)
The haemoglobin of a human foetus

Options:
A has a lower affinity for oxygen than that of the adult and the oxygen dissociation curve is towards right.
B has affinity for oxygen the same as that of an adult.
C has a higher affinity for oxygen than that of the adult and the oxygen dissociation curve is towards right.
D has a higher affinity for oxygen than that of an adult and the oxygen dissociation curve is towards left.

The correct answer is:
has a higher affinity for oxygen than that of an adult and the oxygen dissociation curve is towards left.

Solution:
The haemoglobin of a human foetus has a higher affinity for oxygen than that of an adult. Haemoglobin is a group of globlar proteins occurring widely in animal as oxygen carriers in blood. As a result, the oxygen dissociation curve is towards left.

Topic: Biology, Breathing and Exchange of Gases, NEET
The half life of a first order reaction is 2000 years. If the concentration after 8000 years is $0.02 \mathrm{M}$, then the initial concentration was :

Options:
A $0.16 \mathrm{M}$
B $0.32 \mathrm{M}$
C $0.08 \mathrm{M}$
D $0.04 \mathrm{M}$

The correct answer is:
$0.32 \mathrm{M}$

Solution:
Half life period $\left(t_{1 / 2}\right)=2000$ years
$\begin{aligned}
k & =\frac{0.693}{2000} \\
k & =\frac{2.303}{t} \log \frac{[\mathrm{A}]_o}{[\mathrm{~A}]_t} \\
\frac{0.693}{2000} & =\frac{2.303}{8000} \log \frac{[\mathrm{A}]_o}{0.02} \\
\therefore \quad[\mathrm{A}]_o & =0.32 \mathrm{M}
\end{aligned}$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2022 (Phase 2)
The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to 116th of its initial value?

Options:
A 80 minutes
B 20 minutes
C 40 minutes
D 60 minutes

The correct answer is:
80 minutes

Solution:
Activity of substance is given by, A=A02n, where n is the number of half-lives.

Now, A016=A02n⇒n=4

Therefore,

n=tT12⇒t=4T12=4×20 min=80 min

Topic: Physics, Nuclear Physics, NEET, NEET 2023 (All India)
The half life of radioactive sample undergoing α-decay is 1.4×1017 s. If the number of nuclei in the sample is 2.0×1021, the activity of the sample is nearly:

Options:
A 104 Bq
B 105 Bq
C 106 Bq
D 103 Bq

The correct answer is:
104 Bq

Solution:
Recall the formula of activity of a radioactive element in terms of decay constat, R=λN, and we know that the relation between decay constant and half-life is, T=0.693λ⇒R=0.693T×N⇒R=0.6931.4×1017×2×1021⇒R=104 Bq.

Topic: Physics, Nuclear Physics, NEET, NEET 2020 (Phase 2)
The half life period for catalytic decomposition of XY3 at 100 mm is found to be 8 hrs and at 200 mm it is 4 hrs. The order of reaction is

Options:
A 3
B 2
C 4
D None of these

The correct answer is:
2

Solution:
The value of order of reaction using half life and pressure values can be find out by using this relation given below
t121t122=P2P1n-1
or 84=200100n-1
(2)1=(2)n-1
or 1=n-1
So, n=2.
Hence, it is second order reaction.

Topic: Chemistry, Chemical Kinetics, NEET
The half - life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially, they have the same number of atoms. Then,

Options:
A X will decay faster than Y
B Y will decay faster than X
C X and Y have the same decay rate initially
D X and Y decay at the same rate always

The correct answer is:
Y will decay faster than X

Solution:
Let λX and λY be the decay constants of the radioactive elements X and Y respectively.
Here, half - life period of element X = mean life time of element Y
∴ 0.693λX=1λY or λX=0.693 λY
or λY>λX
Therefore, the element Y will decay faster.

Topic: Physics, Nuclear Physics, NEET
The half-life for a zero order reaction having 0.02 M initial concentration of reactant is 100 s. The rate constant (in mol L-1s-1) for the reaction is

Options:
A 1.0×10-4
B 2.0×10-4
C 2.0×10-3
D 1.0×10-2

The correct answer is:
1.0×10-4

Solution:
t1/2zero =[A]02K

100s=0.02M2K
K=0.02M2×100=1×10-4 mol L-1s-1

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2020 (Phase 2)
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, NEET
The half-life for radioactive decay of14C is 5730 years. An archaeological artifact containing wood had only 80% of the14C found in a living tree. Estimate the age of the sample.

Options:
A 5686 years
B 4444 years
C 3256 years
D 1845 years

The correct answer is:
1845 years

Solution:
Correct Option is : (D)
1845 years

Topic: Chemistry, Chemical Kinetics, NEET
The half-life of 92238U against Î±â decay is 4.5Ã 10 9 years. Calculate the activity of 1 g sample of 92238U.

Options:
A 4.231×104 dps
B 1.721×103 dps
C 1.235×104 dps
D 5.167×103 dps

The correct answer is:
1.235×104 dps

Solution:
T 1 2 =4.5× 10 9 years

=4.5×109×3.156×107 s

No. of atoms in 1 g Uranium,

N= 6.023× 10 23 238 atoms

The activity of the sample =λ N

= 0.693 T 1 2 ×N

= 0.693×6.023× 10 23 4.5×3.156× 10 16 ×238

=1.235× 10 4   distintegration/second.

Topic: Physics, Nuclear Physics, NEET
The half-life of 131 I is 8 days. Given a sample of 131 I at time t=0, we can assert that

Options:
A No nucleus will decay before t=4 days
B No nucleus will decay before t=8 days
C No nucleus will decay before t=16 days
D A given nucleus may decay at any time after t=0

The correct answer is:
A given nucleus may decay at any time after t=0

Solution:
Number of nuclei decreases exponentially
Nâ¡ = Nâ¡ 0 eâ¡ - Î» t and rate of decay - dâ¡ Nâ¡ dâ¡ t = Î» Nâ¡
Therefore, decay process lasts upto t = â.
Therefore, a given nucleus may decay at any time after t = 0

Topic: Physics, Nuclear Physics, NEET
The half-life of a radioactive isotope $X$ is $20 \mathrm{yr}$. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio $1: 7$ in a sample of a given rock. The age of the rock is estimated to be

Options:
A $40 \mathrm{yr}$
B $60 y \mathrm{r}$
C $80 \mathrm{yr}$
D $100 \mathrm{yr}$

The correct answer is:
$60 y \mathrm{r}$

Solution:
$$
\text { ( As } \begin{aligned}
\frac{N}{N_0}= & \left(\frac{1}{2}\right)^n \\
& \frac{N}{N_0}=\left(\frac{1}{2}\right)^3=\frac{1}{8}
\end{aligned}
$$
Number of half lives $=3$
$$
\begin{aligned}
\Rightarrow & T & =20 \mathrm{yr} \\
\therefore & T & =\frac{t}{n}
\end{aligned}
$$
or $\quad t=T \times n=20 \times 3 \mathrm{yr}=60 \mathrm{yr}$

Topic: Physics, Nuclear Physics, NEET, NEET 2013 (All India)
The half-life of a radioactive isotope $X$ is $50 \mathrm{yr}$. 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 $200 \mathrm{yr}$
B $250 \mathrm{yr}$
C $100 \mathrm{yr}$
D $150 \mathrm{yr}$

The correct answer is:
$200 \mathrm{yr}$

Solution:
We know that
$$
\begin{aligned}
\frac{N}{N_0} & =\left(\frac{1}{2}\right)^{t / t_1 / 2} \\
\frac{1}{16} & =\left(\frac{1}{2}\right)^{t / 50} \\
t & =4 \times 50 \\
t & =200 \mathrm{yr}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2011 (Screening)
The half-life of a radioactive sample is T. If the activities of the sample at a time t1 and t2 t1<t2 are R1 and R2 respectively, then the number of atoms disintegrated in time t2-t1 is proportional to

Options:
A R1-R2T
B R1+R2T
C R1R2R1+R2T
D R1+R2T

The correct answer is:
R1-R2T

Solution:
Activity R=Î»N
So that
R1=Î»N1 and R2=Î»N2
R1-R2=Î»N1-N2=0.6931TN1-N2
N1-N2=R1-R2T0.6931
N1-N2 âR1-R2T

Topic: Physics, Nuclear Physics, NEET
The half-life of a radioactive substance is 30 min. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

Options:
A 15
B 30
C 45
D 60

The correct answer is:
60

Solution:
Number of active nuclei falls from 60% to 15%
So the sample becomes 14th
=122th
So the number of half-lives = 2
Time t=2×30=60 min

Topic: Physics, Nuclear Physics, NEET, NEET 2016 (Phase 2)
The half-life of a radioactive substance is 30 min. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

Options:
A 15
B 30
C 45
D 60

The correct answer is:
60

Solution:
Number of active nuclei falls from 60% to 15%
So the sample becomes 14th
=122th
So the number of half-lives = 2
Time t=2×30=60 min

Topic: Physics, Nuclear Physics, NEET
The half-life of a radioactive substance is 50 days. The substance will disintegrate completely in

Options:
A 50 days
B 500 days
C 5000 days
D infinite time

The correct answer is:
infinite time

Solution:
We know that N=N0e-Î»t
N is equal to N0 at t=0
N becomes 0 at t=â,
Initially, the graph falls rapidly but later the rate of decrease reduces with time.

Topic: Physics, Nuclear Physics, NEET
The half-life of At215 is 100 Î¼s. The time taken for the radioactivity of a sample of At215 to decay to 1 1 6 th of its initial value is

Options:
A 400 μs
B 6.3 μs
C 40 μs
D 300 μs

The correct answer is:
400 μs

Solution:
If R is activity of radioactive substance after n half lives then

R=R012n; R016=R012n
∴ n=4; t=nT1/2=4×100=400 μs

Topic: Physics, Nuclear Physics, NEET
The half-life of Radioactive substance is 20 min. The approximate time interval (t2-t1) between the time t2 when 23 of it has decayed and time t1 when 13 of it had decayed is

Options:
A 14 min

B 20 min

C 28 min

D 7 min

The correct answer is:
20 min

Solution:
N1=N0 - 13N0 =23N0
N2=N0 - 23N0 =13N0
â´ N2N1 = 12n
n=1
â´ t2- t1 = one half âlife = 20 min

Topic: Physics, Nuclear Physics, NEET
The half-life of radium is 1600 years. The time does 1g of radium take to reduce to 0.125g is

Options:
A 4800 years
B 4500 years
C 5000 years
D 4750 years

The correct answer is:
4800 years

Solution:
Nuclear reactions are Ist order reaction
NN0=12n

â 0.125 1 = ( 1 2 ) n
= 1 8 = 1 2 n
ân=3
n=tt12 ;
âµââ3= t 1600
t=4800

Topic: Chemistry, Chemical Kinetics, NEET
The half-life of the radioactive nucleus is 50 days. The time interval ( t 2 â t 1 ) between the time t 2 when 2 3 of it has decayed and the time t 1 when 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:
At time t 2, 2 3 of the sample had decayed
â´ Nâ¡ = 1 3 Nâ¡ 0
â´ 1 3 Nâ¡ 0 = Nâ¡ 0 eâ¡ - Î» t 2 .....(i)
At time t 1 , 1 3 of the sample had decayed
â´ Nâ¡ = 2 3 Nâ¡ 0
â´ 2 3 Nâ¡ 0 = Nâ¡ 0 eâ¡ - Î» t 1 .....(ii)
Divide (i) by (ii), we get
1 2 = eâ¡ - Î» t 2 eâ¡ - Î» t 1
1 2 = eâ¡ - Î» t 2 - t 1
Î»( t 2 â t 1 )= ln2
t 2 - t 1 = ln 2 Î» = ln 2 ln 2 Tâ¡ 1 / 2 âµ Î» = ln 2 Tâ¡ 1 / 2
=T 1/2 =50 days

Topic: Physics, Nuclear Physics, NEET
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 [log2=0.301]

Options:
A 23.03 minutes
B 46.06 minutes
C 460.6 minutes
D 230.3 minutes

The correct answer is:
46.06 minutes

Solution:
As k=0.6932t1/2
=0.69326.93 min-1
Also t=2.303Î»log[Ao][A]
[An] = initial concentration (amount)
[A] = final concentration (amount)
So t=2.303Ã6.930.6932log1001
=46.06 minutes

Topic: Chemistry, Chemical Kinetics, NEET
The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially both of them have the same number of atoms. Then,

Options:
A x and y have the same decay rate initially
B x and y decay at the same rate always
C y will decay at a faster rate than x
D x will decay at a faster rate than y

The correct answer is:
y will decay at a faster rate than x

Solution:
t12x = tmeany
or 0 Â· 6 9 3 Î» z = 1 Î» y
Therefore, Î»x<Î»y=0.693 Î»y
Î»x<Î»y
or Rate of decay = Î»N
Initially number of atoms (N) of both are equal but since Î»y>Î»x, therefore, y will decay at a rate faster than x

Topic: Physics, Nuclear Physics, NEET
The half-life period of a radioactive element X is the same as the mean lifetime of another radioactive element Y. Initially, they have the same number of atoms. Then

Options:
A X and Y decay at same rate always
B X will decay faster than Y
C Y will decay faster than X
D X and Y have same decay rate initially

The correct answer is:
X will decay faster than Y

Solution:
According to the question,
T 1 2 ( X )=Ï( Y )
â 0.693Î»X=1Î»Y
â Î»Y=Î»X0.693
â Î»Y>Î»X
This means, X will decay faster than Y.

Topic: Physics, Nuclear Physics, NEET
The halide which is inert to water is

Options:
A PCl5
B SiCl4
C BCl3
D SF6

The correct answer is:
SF6

Solution:
Sometimes, when maximum covalency is obtained, the halides become inert to water. Thus, SF6 (or similarly CCl4 ) is stable. This is because SF6 is coordinately saturated and sterically hindered. Thus, SF6 is inert to water, because of kinetic rather than thermodynamic factor.
A. PCl5+4H2O→H3PO4+5H+Cl-
B. SiCl4+4H2O→Si(OH)4+4H+Cl-
C. BCl3+3H2O→B(OH)3+3H+Cl-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?

Options:
A With higher oxidation states, the transition metals have a larger number of unpaired electrons.
B With higher oxidation states, the transition metals have a smaller atomic size.
C With higher oxidation states, the transition metals have a higher charge density.
D With higher oxidation states, the transition metals have a lower ionization energy.

The correct answer is:
With higher oxidation states, the transition metals have a higher charge density.

Solution:
Correct Option is : (C)
With higher oxidation states, the transition metals have a higher charge density.

Topic: Chemistry, d and f Block Elements, NEET