datavorous/entrance-exam-dataset huggingface dataset:
The element which helps in oxygen evolution in the process of photosynthesis is

Options:
A Zn and Mn
B Mo and Cl
C B and Mg
D Cl and Mn

The correct answer is:
Cl and Mn

Solution:
Manganese and chlorine are essential co-factor for photosynthesis. It also functions in the photolysis of water, which is responsible for the evolution of oxygen.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The element which prevents bursting of the pollen tube and helps in pollen germination is

Options:
A K
B B
C Fe
D H2O

The correct answer is:
B

Solution:
Boron has a regulatory role in pollen germination and pollen tube growth. Boron affected the frequency of pollen germination and the morphology of the pollen tube greatly.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The element with Z = 114 has been discovered recently. It will belong to which of the following family/group in periodic table and its electronic configuration is?

Options:
A Carbon family, [Rn]5f14,6d10,7s27p2
B Oxygen family, [Rn]5f14,6d10,7s27p4
C Nitrogen family, [Rn]5f14,6d10,7s27p6
D Halogen family, [Rn]5f14,6d10,7s27p5

The correct answer is:
Carbon family, [Rn]5f14,6d10,7s27p2

Solution:
Z = 114
Electronic configuration of Z=[Rn]5f14,6d10,7s27p2
Period =7 or VII
Block = p
Gp. No =2+2= IV (carbon family)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The element Z=114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?

Options:
A Halogen family, Rn 5f14 6d10 7s2 7p5
B Carbon family, Rn 5f14 6d10 7s2 7p2
C Oxygen family, Rn 5f14 6d10 7s2 p4
D Nitrogen family, Rn5f14 6d10 7s27p3

The correct answer is:
Carbon family, Rn 5f14 6d10 7s2 7p2

Solution:
Z=114 belong to group 14, carbon family.
Electronic configuration =Rn5f14 6d10 7s2 7p2

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2017
The elemental phosphorus is made from rock phosphate, Ca3(PO4)2 by making use of which one of the following reactions?

Options:
A Ca3(PO4)3+C+MgO→Δ
B Ca3(PO4)2+C+SiO2→Δ
C Ca3(PO4)2+C+ZnO→Δ
D Ca3(PO4)2+C+FeO→Δ

The correct answer is:
Ca3(PO4)2+C+SiO2→Δ

Solution:
Ca3(PO4)2+SiO2→ΔCaSiO3+P2O5
P2O5+C→ΔP+CO

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The elementary sulphur is present in

Options:
A Valine, lysine and cystine
B Tryptophan, Glutamic acid, Aspartic acid
C Citrulline, Methionine and Glutamic acid
D Homo-cysteine, Cystine and Methionine

The correct answer is:
Homo-cysteine, Cystine and Methionine

Solution:
Amino acids are amphoteric in nature both acidic and alkyl group present. Sulphur containing amino acids are cysteine, methionine and cystine is non-protein amino acids

Topic: Chemistry, Biomolecules, NEET
The elements having the maximum and minimum melting points among the members of group 13 are respectively -

Options:
A B, Tl
B Tl, B
C B, Ga
D Al, Tl

The correct answer is:
B, Ga

Solution:
Melting point in group - 13 decreases down the group but the trend is not regular. Boron has the highest melting point because of its very strong covalent structure. Gallium has a very low melting point (29.8 K) and, therefore, could exist in liquid state during summer.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The elements Z = 117 have not yet been discovered. In which family/group would you place this element, and also give its electronic configuration?

Options:
A Halogens, [Rn] 5f146d107s27p5
B Noble Gases, [Xe] 4f145d106s26p5
C Halogens, [Kr] 4d105s25p5
D Alkali Metals, [Ne] 3s23p6

The correct answer is:
Halogens, [Rn] 5f146d107s27p5

Solution:
Correct Option is : (A)
Halogens, [Rn] 5f146d107s27p5

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The elements Z = 120 have not yet been discovered. In which family/group would you place this element, and also give its electronic configuration?

Options:
A Alkaline Earth Metals, [Xe] 4f145d106s26p67s2
B Noble Gases, [Kr] 4d105s25p66s26p6
C Alkaline Earth Metals, [Uuo] 8s2
D Alkali Metals, [Ne] 3s23p64s24p65s2

The correct answer is:
Alkaline Earth Metals, [Uuo] 8s2

Solution:
Correct Option is : (C)
Alkaline Earth Metals, [Uuo] 8s2

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The E.M. wave with shortest wavelength among the following is

Options:
A Ultraviolet rays
B X-rays
C Gamma-rays
D Microwaves

The correct answer is:
Gamma-rays

Solution:
If we look at the spectrum of electromagnetic waves then the order of a wavelength is, λgamma<λX-ray<λUV rays<λVisible<λInfrared<λMicrowave<λRadio, so from the above spectrum the wavelength of gamma rays is minimum and then the wavelength of a radio wave is maximum, so the shortest wavelength is for gamma-rays.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2020 (Phase 2)
The embedding of _________X________ in the _________Y_________of uterus is called implantation. Select the correct option for X and Y.

Options:
A

X	Y
Blastocyst	Endometrium

B

X	Y
Blastocyst	Myometrium

C

X	Y
Trophoblasts	Myometrium

D

X	Y
Trophoblasts	endometrium

The correct answer is:

X	Y
Blastocyst	Endometrium

Solution:
The embedding of Blastocysts in the endometrium of uterus in the presence of progesterone hormone is called implantation. It occurs after 7th day of fertilization. Blastocyst is a 64-celled stage with a blastocoel cavity.

Topic: Biology, Human Reproduction, NEET
The embryogenesis is the process of the development of an embryo from the zygote. During this process zygote undergoes

Options:
A Meiosis
B Cleavage only
C Cell differentiation only
D Both cleavage and cell differentiation

The correct answer is:
Both cleavage and cell differentiation

Solution:
Embryogenesis is a developmental process that usually begins once the egg has been fertilized. It involves the multiplication of cells by cleavage and their subsequent differentiation into tissues and organs of a living organism.

Topic: Biology, Human Reproduction, NEET
The EMF of a cell corresponding to the reaction :

Zn(s)+2 H + (aq)→Z n 2+ (0.1  M)+ H 2 (g)  (1  atm.)  is  0.28  volt  at  25°C.

The pH of the solution at the hydrogen electrode.

E Z n 2+  / Zn o =−0.76  volt;   E H +  /  H 2 o =0

Options:
A

2.30

B

7.8

C

9.2

D

8.30

The correct answer is:

8.30

Solution:
The cell may be represented as Zn(s)| Zp 2 ( aq ) || H( aq ) | H 2 (g)1atmPt(s)
Zn( s )            →      Zn 2+ ( aq )+2 e ¯  ( Anode ) 2 H + ( aq )+2 e ¯   →       H 2 (g)  1atm             ( Cathode )
----------------------------------------------------------
Zn( s ) +2H + ( aq ) → 2 e - Zn 2+ ( aq ) +H 2 (g) 1atm (CellReaction)
E cell = 0 . 2 8 = E cell o - 0 . 0 5 9 2 log Zn aq 2 + H + 2
0 . 2 8 = 0 - - 0 . 7 6 - 0 . 0 5 9 2 log 0 . 1 H + 2
0 . 2 8 - 0 . 7 6 = - 0 . 4 8 = - 0 . 0 5 9 2 log 0 . 1 H + 2
0 . 4 8 Ã 2 0 . 0 5 9 = 1 6 . 2 7 = log 0 . 1 H + 2
0 . 1 H + 2 = 1 . 8 6 2 Ã 1 0 + 1 6
H + 2 = 0 . 1 1 . 8 6 2 Ã 1 0 + 1 6
=0.0537Ã 10 â16/2
H + = 5 . 3 7 Ã 1 0 - 9
pH = - log (5.37 x 10 -9)
= 9 - log 5,37
= 9 - 0.73 = 8.27
â 8 . 3

Topic: Chemistry, Electrochemistry, NEET
The emf of a cell having internal resistance $1 \Omega$ is balanced against a length of $330 \mathrm{~cm}$ on a potentiometer wire. When an external resistance of $2 \Omega$ is connected across the cell, the balancing length will be

Options:
A $115 \mathrm{~cm}$
B $332 \mathrm{~cm}$
C $220 \mathrm{~cm}$
D $330 \mathrm{~cm}$

The correct answer is:
$220 \mathrm{~cm}$

Solution:
For case - I
$\begin{aligned} & \Rightarrow \varepsilon_s=\phi \times 330 \\ & \Rightarrow \phi=\frac{\varepsilon_s}{330}\left(\frac{V}{\mathrm{~cm}}\right) \ldots\end{aligned}$
For case - II
$\begin{aligned} & \Rightarrow i \times 2=\phi I_2 \\ & \Rightarrow \frac{\varepsilon_s}{3} \times 2=\frac{\varepsilon_s}{330} \times I_2 \text { [From equation 1] } \\ & \Rightarrow I_2=220 \mathrm{~cm}\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2023 (Manipur)
The emf of a cell is 6 V and internal resistance is 0.5 kÎ©. The reading of a voltmeter having an internal resistance of 2.5 kÎ© is

Options:
A 6 V
B 10Â V
C 5 V
D 0.5Â V

The correct answer is:
5 V

Solution:
Total resistance=2.5+0.5=3.0 kÎ©=3000 Î©
â´ Current, i=63000 A;
Reading of voltmeter =iÃ2.5Ã1000
=63000Ã2500=5 V

Topic: Physics, Current Electricity, NEET
The e.m.f. of a Deniell cell at $298 \mathrm{~K}$ is $\mathrm{E}_1$
$$
\mathrm{Zn}\left|\begin{array}{c}
\mathrm{ZnSO}_4 \\
(0.01 \mathrm{M})
\end{array}\right|\left|\begin{array}{c}
\mathrm{CuSO}_4 \\
(0.01 \mathrm{M})
\end{array}\right| \mathrm{Cu}
$$
When the concentration of $\mathrm{ZnSO}_4$ in 1.0 $\mathrm{M}$ and that of $\mathrm{CuSO}_4$ is $0.01 \mathrm{M}$, the e.m.f. changed to $\mathrm{E}_2$. What is the relationship between $E_1$ and $E_2$ ?

Options:
A $\mathrm{E}_1>\mathrm{E}_2$
B $\mathrm{E}_1 < \mathrm{E}_2$
C $\mathrm{E}_1=\mathrm{E}_2$
D $\mathrm{E}_2=0 \neq \mathrm{E}_1$

The correct answer is:
$\mathrm{E}_1>\mathrm{E}_2$

Solution:
Cell reaction can be represented as:
$$
\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Ca}+\mathrm{Zn}^{2+}
$$
Applying in both cases,
$$
\mathrm{E}^0=\frac{-0.0591}{2} \log \frac{\mathrm{Zn}^{2+}}{\mathrm{C} \mu^{2+}}
$$
Thus, $\mathrm{E}_1>\mathrm{E}_2$

Topic: Chemistry, Electrochemistry, NEET, NEET 2003
The emf of the following three galvanic cells.
I. Zn/Zn2+(1M)||Cu2+(1M)/Cu
II. Zn/Zn2+(0.1M)||Cu2+(1M)/Cu
III. Zn/Zn2+(1M)||Cu2+(0.1M)/Cu
are represented by E1,E2 and E3 . Which of the following is true

Options:
A E1>E2>E3
B E3>E1>E2
C E3>E2>E1
D E2>E1>E3

The correct answer is:
E2>E1>E3

Solution:
Ecell=Ecello-2.303RTnFlog[Zn2+][Cu2+]
E1=Ecello-2.303RT2Flog11=Ecello
E2=Ecello-2.303RT2Flog0.11
=Ecello+2.303RT2F
E2=Ecello-2.303RT2Flog10.1
=Ecello-2.303RT2F
Hence, E2>E1>E3 .

Topic: Chemistry, Electrochemistry, NEET
The emitter-base junction of a transistor is ..........biased while the collector base junction is ........biased

Options:
A Forward, Forward
B Forward, Reverse
C Reverse, Forward
D Reverse, Reverse

The correct answer is:
Forward, Reverse

Solution:
If emitter is P type, it is connected to positive plate of battery, and base N type is connected to negative plate of battery, hence we can say that emitter base is forward biased
Similarly if collector is P type, it is connected to negative plate of battery, and base N type is connected to positive plate of battery, hence we can say that collector base is reversed biased

Topic: Physics, Semiconductors, NEET
The empirical formula of a mixed oxide, in which the oxide ions are present in the CCP lattice positions, half of the octahedral voids are occupied by trivalent ions Y3+ and one-fifth of tetrahedral voids are occupied by divalent X2+ ions, will be

Options:
A XY2O4.
B X2YO4.
C X4Y5O10.
D X5Y4O10.

The correct answer is:
X4Y5O10.

Solution:
In CCP, anions occupy the face-centered and the corner positions of the cube.

Based on the position and number of voids in CCP, for one oxide ion there are two tetrahedral voids and one octahedral void.
Since one-fifth of the tetrahedral voids are occupied by divalent cations X2+-
Number of divalent cations for each oxide ion=2×15
Also, since half of the octahedral voids are occupied by trivalent cations Y3+-
Number of trivalent cations for each oxide ion=1×12

The empirical formula gives the simplest ratio of all the atoms present in a molecule. So, the formula of the compound is X25Y12O or X4Y5O10

Topic: Chemistry, Solid State, NEET
The end correction of a resonance column is 1 cm. If the shortest length resonating with the tuning fork is 10 cm, the next resonating length should be

Options:
A 32 cm
B 40 cm
C 28 cm
D 36 cm

The correct answer is:
32 cm

Solution:
Given: e=1 cm
For first resonance,
Î»4= l1+e=11 cm
For second resonance,
3Î»4= l2+e â l2=3 Ã11-1=32 cm

Topic: Physics, Waves and Sound, NEET
The end of juvenile phase mark the beginning of

Options:
A Vegetative phase
B Reproductive phase
C Senescence
D Ageing

The correct answer is:
Reproductive phase

Solution:
The life span is divided into the juvenile phase, reproductive phase, and senescence.

Topic: Biology, Reproduction in Organisms, NEET
The end of the T-wave in ECG marks the

Options:
A beginning of ventricular systole
B end of ventricular systole
C beginning of atrial diastole
D end of atrial diastole

The correct answer is:
end of ventricular systole

Solution:
ECG is an electrocardiogram that is a graph that represents heart activity. This is recorded by a machine known as an electrocardiograph. This is recorded using six chest leads and three limb leads. This graph shows many peaks that are designated by alphabets P to T. P marks the beginning and T is the end of the graph. Heart activity begins with the help of sino atrial node and this is represented by a P wave. Hence P wave denotes atrial depolarization. QRS represents ventricular systole and the ending of this is represented by T wave.

Topic: Biology, Body Fluids and Circulation, NEET
The end product of fermentation of molasses by yeast is

Options:
A Acetic acid
B Methyl alcohol
C Ethyl alcohol
D Lactate

The correct answer is:
Ethyl alcohol

Solution:
In fermentation, say by yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to CO2 and ethanol. The enzymes, pyruvic acid decarboxylase and alcohol dehydrogenase catalyze these reactions.

Topic: Biology, Microbes in Human Welfare, NEET
The end product of ornithine cycle is

Options:
A Ammonia
B Urea
C Arginine
D Carbon dioxide

The correct answer is:
Urea

Solution:
The entire process of this cycle converts two amino groups, one from NH4- and one from Asp, and a carbon atom from HCO3-, to the relatively nontoxic excretion product urea at the cost of four "high-energy" phosphate bonds. To enter the cycle, ammonia is converted to carbomyl phosphate.

Topic: Biology, Excretory Products and their Elimination, NEET
The end product of the Î²- decay of 1532P is

Options:
A S1632
B S1635
C N1432
D Cl1732

The correct answer is:
S1632

Solution:
The following equation shows the Î²- decay of 1532P

1532Pâ1632S+Î²â

1532P â 1632S +e-10

Topic: Physics, Nuclear Physics, NEET
The end product of the decay of 90Th232 is 82Pb208. The number of Î± and Î²-particles emitted are respectively

Options:
A 6, 4
B 3, 3
C 4, 6
D 6, 6

The correct answer is:
6, 4

Solution:
Let α-particles emitted are x and β-particles emitted are y
90Th232 → 82Pb208 + x 2He4 + y -1e0
On comparing the atomic number
90=82+2x-y
or 2x-y=8 ...(i)
On comparing the mass number
232=208 + 4x
or x=6
Putting the value of x in Eq.(i), we get
y=4

Topic: Physics, Nuclear Physics, NEET
The end product of the following reaction would be

Options:
A
B
C
D

The correct answer is:

Solution:
Hydrolysis of ether linkages and acyl-oxygen cleavage by acidic hydrolysis form beta keto acid which on heating undergoes further decarboxylation leads to cyclohexanone.

Topic: Chemistry, Aldehydes and Ketones, NEET
The endocrine gland that produces somatostatin also produces

Options:
A Insulin and glucagon
B Progesterone and oestrogen
C Thyroxine and calcitonin
D Somatotropin and prolactin

The correct answer is:
Insulin and glucagon

Solution:
Progesterone and Oestrogen are secreted by ovaries and play a major role in the development of the female reproductive system. Thyroxine and calcitonin are secreted by the thyroid gland. Thyroxine plays an important role in the regulation of various organs of the body whereas calcitonin plays a major role in blood calcium level regulation. Somatotropin (growth hormone) and prolactin are secreted by the anterior pituitary gland. Insulin and Glucagon are secreted by the pancreas and play a major role in the regulation of blood glucose levels. Somatostatin is also secreted by the pancreas (delta cells of islets of Langerhans) has acted as a regulating hormone that regulates the levels of both insulin and glucagon. It is also secreted by the hypothalamus.

Topic: Biology, Chemical Coordination and Integration, NEET
The endodermis is a part of:

Options:
A Cortex
B Pericycle
C Medulla
D Epidermis

The correct answer is:
Cortex

Solution:
Cortex includes hypodermis, general cortex and endodermis. Endodermis of roots consists of casparian strips and endodermis of stem consists of the starch sheath.

Topic: Biology, Anatomy of Flowering Plants, NEET
The endoplasmic reticulum is closely associated with the Golgi apparatus. Synthesized proteins from the RER are released from which part of the Golgi apparatus?

Options:
A Cis face of Golgi apparatus
B Cisternae
C Trans face of Golgi apparatus
D Both (a) and (c)

The correct answer is:
Trans face of Golgi apparatus

Solution:
A number of proteins synthesised by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus before they are released from its trans face.

Topic: Biology, Cell - The Unit of Life, NEET
The endosperm, a product of double fertilization in angiosperms is absent in the seeds of:

Options:
A Wheat
B Groundnut
C Maize
D Castor

The correct answer is:
Groundnut

Solution:
The endosperm is the nutritive tissue present in seed which provides nutrition to the developing embryo. In some seeds, endosperm is completely consumed during development of embryo, thus mature seeds lacks endosperm, such seeds are called as non-endospermic seeds, e.g., pea, groundnut.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The endosperm is completely consumed by the developing embryo in the seeds of:

Options:
A Maize and rice
B Rice and castor
C Castor and maize
D Pea and beans

The correct answer is:
Pea and beans

Solution:
Non-albuminous seeds have no residual endosperm as it is completely consumed during embryo development (e.g., pea, groundnut). Albuminous seeds retain a part of endosperm as it is not completely used up during embryo development (e.g., wheat, maize, barley, castor).

Topic: Biology, Morphology of Flowering Plants, NEET
The 'endospores' are formed in the following genera

Options:
A Bacillus and Clostridium
B Mucor and Bacillus
C Monococcus and Clostridium
D Saccharomyces and Clostridium

The correct answer is:
Bacillus and Clostridium

Solution:
An endospore is a dormant, tough, and non-reproductive internal structure produced by some bacteria and archaebacteria. Endospore formation is more common in rod shaped bacteria or bacillus forms.

Topic: Biology, Biological Classification, NEET
The ends of a rod of length l and mass m are attached to two identical springs as shown in the figure. The rod is free to rotate about its centre O. The rod is depressed slightly at end A and released. The time period of the oscillation is

Options:
A 2πm2k
B 2π2mk
C π2m3k
D π3m2k

The correct answer is:
π2m3k

Solution:
Let the rod be depressed by a small amount x (in the figure). Both the springs are compressed by x. When the rod is released, the restoring torque is given by
Ï=kxÃl2+kxÃ12=kxl
Now tanÎ¸=xl/2=2xl.
Since Î¸ is small, tanÎ¸âÎ¸, where Î¸ is expressed in radian.
Thus Î¸=2xl or x=Î¸l2
Ï=kÎ¸l2Ãl=kÎ¸l22
If I is the moment of inertia of the rod about O, then
Id2Î¸dt2=-kl22Î¸
Or d2Î¸dt2=-kl22IÎ¸
Since d2Î¸dt2â-Î¸, the motion is simple harmonic whose angular frequency is given by
Ï=kl22I
Now, Ï=2ÏT and I=ml212. Therefore, we have
2ÏT=kl22Ã12ml2=6km
Or T=Ï2m3k, which is choice Ï2m3k

Topic: Physics, Oscillations, NEET
The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment the displacement of the wire is y 1 = A sin Ï x L sin Ï t and energy is E1 and in other experiment its displacement is y 2 = A sin 2Ï x L sin 2Ï t and energy is E2. Then

Options:
A E2=E1
B E2=2E1
C E2=4E1
D E2=16E1

The correct answer is:
E2=4E1

Solution:
Energy E∝A2f2

Here Amplitude (A) is the same in both the cases, but the frequency in the second case is two times the frequency in the first case,
Therefore, E2=4E1

Topic: Physics, Waves and Sound, NEET
The energies $E_1$ and $E_2$ of two radiations are $25 \mathrm{eV}$ and $50 \mathrm{eV}$ respectively. The relation between their wavelengths i.e., $\lambda_1$ and $\lambda_2$ will be

Options:
A $\lambda_1=2 \lambda_2$
B $\lambda_1=4 \lambda_2$
C $\lambda_1=\frac{1}{2} \lambda_2$
D $\lambda_1=\lambda_2$

The correct answer is:
$\lambda_1=2 \lambda_2$

Solution:
$$
\begin{aligned}
& E_1=25 \mathrm{eV}, E_2=50 \mathrm{eV} \\
& E_1=\frac{h c}{\lambda_1} \text { and } E_2=\frac{h c}{\lambda_2} \\
& \text { or } \frac{E_1}{E_2}=\frac{\lambda_2}{\lambda_1}
\end{aligned}
$$
or $\frac{25}{50}=\frac{\lambda_2}{\lambda_1}$
or $\lambda_1=2 \lambda_2$

Topic: Chemistry, Structure of Atom, NEET, NEET 2011 (Screening)
The energies of orbitals of hydrogen atom are in the order

Options:
A 3s < 3p < 4s < 3d < 4p
B 3s < 3p < 3d < 4s < 4p
C 3s = 3p = 3d < 4s = 4p
D 3s = 3p = 3d < 4s < 4p

The correct answer is:
3s = 3p = 3d < 4s = 4p

Solution:
In hydrogen atom, the energy of electron depends upon principle quantum number only and not on azimuthal quantum number due to lack of inter electronic repulsions.

Topic: Chemistry, Structure of Atom, NEET
The energy absorbed by each molecule $\left(A_2\right)$ of a substance is $4.4 \times 10^{-19} \mathrm{~J}$ and bond energy per molecule is $4.0 \times 10^{-19} \mathrm{~J}$. The kinetic energy of the molecule per atom will be

Options:
A $2.0 \times 10^{-20} \mathrm{~J}$
B $2.2 \times 10^{-19} \mathrm{~J}$
C $2.0 \times 10^{-19} \mathrm{~J}$
D $4.0 \times 10^{-20} \mathrm{~J}$

The correct answer is:
$2.0 \times 10^{-20} \mathrm{~J}$

Solution:
$K E$ of molecule $=$ energy absorbed by molecule
- BE per molecule
$$
\begin{aligned}
& =\left[\left(4.4 \times 10^{-19}\right)-\left(4.0 \times 10^{-19}\right) \mathrm{J}\right. \\
& =0.4 \times 10^{-19} \mathrm{~J} \\
& \text { KE per atom }=\frac{0.4 \times 10^{-19}}{2} \mathrm{~J}=2.0 \times 10^{-20} \mathrm{~J}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2009 (Screening)
The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18J atom–1. What is the energy associated with the fifth orbit?

Options:
A -7.72 ×10−20J/atom
B -5.72 ×10−20J/atom
C -3.72 ×10−20J/atom
D -8.72 ×10−20J/atom

The correct answer is:
-8.72 ×10−20J/atom

Solution:
Correct Option is : (D)
-8.72 ×10−20J/atom

Topic: Chemistry, Structure of Atom, NEET
The energy band gap is maximum in

Options:
A metals
B superconductors
C insulators
D semiconductors

The correct answer is:
insulators

Solution:
The energy band gap is maximum in insulators.

Topic: Physics, Semiconductors, NEET
The energy equivalent of 0.5 g of a substance is:

Options:
A 4.5×1013 J
B 1.5×1013 J
C 0.5×1013 J
D 4.5×1016 J

The correct answer is:
4.5×1013 J

Solution:
Einstein's theory of special relativity gives the relation between mass and energy.

Energy, E=mc2, where c is the speed of light.

E=0.5×10−3×9×1016

⇒E=4.5×1013 J

Topic: Physics, Nuclear Physics, NEET, NEET 2020 (Phase 1)
The energy equivalent to a substance of mass 1 g is

Options:
A 18Ã1013 J
B 9Ã1013Â JÂ
C Â 18Ã106Â JÂ
D 9Ã106Â JÂ

The correct answer is:
9×1013 J

Solution:
Given, m=1 g=10-3 kg
Energy, E=mc2
â´ Energy equivalent =10-3Ã3Ã1082
âµc=3Ã108 m s-1
=9Ã1013 J

Topic: Physics, Nuclear Physics, NEET
The energy flux of sunlight reaching the surface of the earth is 1.388 Ã103 W m-2. The number of photons (nearly) per square meter are incident on the earth per second is (Assume that the photons in the sunlight have an average wavelength of 550 nm).

Options:
A 3.838 ×1021 photon m-2 s-1

B 3.838 ×1023 photon m-2 s-1

C 5.838 ×1021 photon m-2 s-1

D 5.838 ×1023 photon m-2 s-1

The correct answer is:
3.838 ×1021 photon m-2 s-1

Solution:
Given, energy per unit area per second, p=1.388×103 W m-2
Let n be the number of photons incident on the earth per square metre. The wavelength of each photon = 550×10-9 m
The energy of each photon, E=hcλ

= 6 . 6 3 × 1 0 - 3 4 × 3 × 1 0 8 5 5 0 × 1 0 - 9 = 3 . 6 1 6 × 1 0 - 1 9 J
Number of photons incident on the earth's surface
n=PE=1.388×1033.616×10-19=3.838×1021
=3.838×1021 photon m-2 s-1

Topic: Physics, Dual Nature of Matter, NEET