datavorous/entrance-exam-dataset huggingface dataset:
The energy band gap is maximum in

Options:
A
metals

B
superconductors

C
insulators

D
semiconductors

The correct answer is:
insulators

Solution:
Maximum in insulators and overlapping in metals

Topic: Physics, Semiconductors, JEE Main, JEE Main 2002
The energy band gap is maximum in

Options:
A metals
B superconductors
C insulators
D semiconductors

The correct answer is:
insulators

Solution:
In insulators, the electrons are tightly bound with the nucleus hence, at room temperature thermal energy is not enough to push the electrons into conduction band and hence, no electrons are available for conduction. The energy required for electron to escape out from orbit and to over come the energy gap is thus of the order of $6 \mathrm{eV}$, while for semiconductors and conductors it is (of the order of) $1 \mathrm{eV}$ and 0 respectively.

Topic: Physics, Semiconductors, JEE Main
The energy band gap of semiconducting material to produce violet (wavelength =4000Å) LED is _____ eV. (Round off to the nearest integer).

Options:

The correct answer is:
3

Solution:
Energy gap of semiconducting material Eg=hcλ=1242λnm=1242400=3.105

Answer rounded to 3eV.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The energy density associated with electric field E→ and magnetic field B→ of an electromagnetic wave in free space is given by (ε0- permittivity of free space, μ0- permeability of free space)

Options:
A UE=E22ε0, UB=B22μ0
B UE=ε0E22, UB=B22μ0
C UE=ε0E22, UB=μ0B22
D UE=E22ε0, UB=μ0B22

The correct answer is:
UE=ε0E22, UB=B22μ0

Solution:
When an electromagnetic wave propagates from the source, it transfers energy to the objects in its path. Electric and magnetic fields serve as energy reservoirs for electromagnetic waves. An electromagnetic wave contains the same amount of energy as are contained in the electric and magnetic fields together. In such situation, the energy density of the electromagnetic wave is equal to the sum of the energies of the electric and magnetic fields.

The average electric and magnetic energy densities are given by

UE=12ε0E2UB=12B2μ0

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The energy dissipated by a resistor is 10 mJ in 1 s, when an electric current of 2 mA flows through it. The resistance is ________Ω. (Round off to the Nearest Integer)

Options:

The correct answer is:
2500

Solution:
Q=i2RT

R=Qi2t=10×10-34×10-6×1=2500 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The energy equivalent of $1 \mathrm{~g}$ of substance is :

Options:
A $5.6 \times 10^{12} \mathrm{MeV}$
B $5.6 \times 10^{26} \mathrm{MeV}$
C $11.2 \times 10^{24} \mathrm{MeV}$
D $5.6 \mathrm{eV}$

The correct answer is:
$5.6 \times 10^{26} \mathrm{MeV}$

Solution:
$\mathrm{E}=\mathrm{mC}^2$
$\begin{aligned} & \Rightarrow \mathrm{E}=\left(1 \times 10^{-3}\right) \times\left(3 \times 10^8\right)^2 \mathrm{~J} \\ & \Rightarrow \mathrm{E}=\left(10^{-3}\right)\left(9 \times 10^{16}\right)\left(6.241 \times 10^{18}\right) \mathrm{eV} \\ & \mathrm{E}=56.169 \times 10^{31} \mathrm{eV} \\ & \mathrm{E} \approx 5.6 \times 10^{26} \mathrm{MeV} \\ & \end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The energy equivalent to a mass of $1 \mathrm{~kg}$ is

Options:
A $9 \times 10^{13} \mathrm{~J}$
B $9 \times 10^{9} \mathrm{~J}$
C $9 \times 10^{16} \mathrm{~J}$
D $9 \times 10^{6} \mathrm{~J}$

The correct answer is:
$9 \times 10^{16} \mathrm{~J}$

Solution:
Energy equivalent of $1 \mathrm{~kg}$ of mass, $\mathrm{E}=\mathrm{mc}^2$
$\begin{aligned}
& =1 \times\left(3 \times 10^8\right)^2 \\
& =9 \times 10^{16} \mathrm{~J}
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
The energy levels of an atom is shown is figure.

Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?

Given (h = 6.62 × 10–34 J s)

Options:
A B
B A
C C
D D

The correct answer is:
D

Solution:
The change in energy is related to wavelength of photon emitted as:

∆E=hcλ=6.62×10-34 J s×3×108 m s-11.6×10-19 J×124.1 nm=1242 eV nm124.1 nm=10 eV

Hence, option D is correct.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is

Options:
A D
B A
C B
D C

The correct answer is:
D

Solution:
Wavelength of photons released during de-excitation is given by,

1λ=R1n12-1n22.

Clearly, shortest wavelength will correspond to highest energy difference in de-excitation, which in the given figure will correspond to transition from n=3 to n=1.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The energy levels of the hydrogen spectrum is shown in the figure. There are some transitions \( A, B, C, D \) and \( E \). Transition \( A, B \) and \( C \) respectively represent

Options:
A First member of Lyman series, third spectral line of Balmer series and the second spectral line of Paschen series.
B Ionisation potential of hydrogen, second spectral line of Balmer series and third spectral line of Paschen series.
C Series limit of Lyman series, third spectral line of Balmer series and second spectral line of Paschen series.
D Series limit of Lyman series, second spectral line of Balmer series and third spectral line of Paschen series.

The correct answer is:
Series limit of Lyman series, third spectral line of Balmer series and second spectral line of Paschen series.

Solution:
Whenever transition is from n=∞, it is called the series limit.

In the transition A from n=∞ to n=1. So, series limit and here, it is to n=1. So, part of the Lyman series.

In transition B, the transition is till n=2. So, part of the Balmer series and is from n=5. So, third spectral line of its Balmer series.

From n=3 to n=2⇒First spectral line of Balmer series.

n=4 to n=2⇒Second spectral line of Balmer series.

So, n=5 to n=2⇒Third spectral line of Balmer series

In transition C, the transition is till n=3, part of the Paschen series. As the transition is from n=5. So, second spectral line of the Paschen series.

Topic: Physics, Atomic Physics, JEE Main
The energy obtained by accelerating an electron by a potential difference of $1 \mathrm{~V}$

Options:
A $1 \mathrm{eV}$
B $2 \mathrm{~J}$
C $1 \mathrm{~W}$
D $1$ $erg$

The correct answer is:
$1 \mathrm{eV}$

Solution:
The energy which an electron acquires when accelerated, through a potential difference of 11 volt is called $1 \mathrm{eV}$. It is a unit of energy. It is usually used as a measure of particle energies although it is not an SI (System International) unit. The SI unit for energy is the Joule.
$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{Joules}$

Topic: Physics, Dual Nature of Matter, JEE Main
The energy of a charged capacitor is given by the expression ( ${ }^q=$ charge on the conductor and $C=$ its capacity)

Options:
A $\frac{q^2}{2c}$
B $\frac{q^2}{C}$
C $2 q C$
D $\frac{q}{2 C^2}$

The correct answer is:
$\frac{q^2}{2c}$

Solution:
$q=C V$ and $U=1 / 2 C V^{\wedge} 2=q^{\wedge} 2 / 2 C$

Topic: Physics, Capacitance, JEE Main
The energy of a photon in a monochromatic light of wavelength $621 \mathrm{~nm}$ matches with the band gap of a semiconducting material. Then the minimum energy required to create an electron-hole pair from the semiconductor is
[Take hc $=1242 \mathrm{eV}-\mathrm{nm}$, where h is Planck's constant and $\mathrm{c}$ is speed of light in vacuum]

Options:
A $3.4 \mathrm{eV}$
B $1.7 \mathrm{eV}$
C $2 \mathrm{eV}$
D $2.2 \mathrm{eV}$

The correct answer is:
$2 \mathrm{eV}$

Solution:
Minimum energy required $=\frac{1242}{621} \mathrm{eV}=2 \mathrm{eV}$

Topic: Physics, Semiconductors, JEE Main
The energy of a photon is calculated by

Options:
A $E=h v$
B $h=E_v$
C $h=\frac{E}{v}$
D $E=\frac{h}{v}$

The correct answer is:
$E=h v$

Solution:
According to the planck's law that energy of a photon is directly proportional to its frequency i.e. $E=h v$

Topic: Chemistry, Structure of Atom, JEE Main
The energy of a photon is E which is equal to the kinetic energy of a proton. If Î»1 be the de-Broglie wavelength of the proton and Î»2 be the wavelength of the photon, then the ratio Î»1Î»2 is proportional to

Options:
A E0
B E12
C E-1
D E-2

The correct answer is:
E12

Solution:
λ1λ2=h2mEhcE⇒λ1λ2=E12

Topic: Physics, Dual Nature of Matter, JEE Main
The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of a proton, $\lambda_2$ the wavelength associated with the proton and if the energy of the photon is $E$, then $\left(\lambda_1 / \lambda_2\right)$ is proportional to

Options:
A $E^4$
B $E^{1 / 2}$
C $E^2$
D $E$

The correct answer is:
$E^{1 / 2}$

Solution:
We know that, de-Broglie wavelength of photon
$$
E=h \mathrm{v}
$$
But we know that also
$$
\begin{aligned}
E & =\frac{1}{2} m v^2 \\
\therefore \quad 2 E & =\frac{P^2}{m} \\
\because \quad P & =m v \\
\frac{P^2}{m} & =m v^2 \\
P & =\sqrt{2 m E}
\end{aligned}
$$
(for proton)
According to the question
$$
\lambda_1=\frac{h}{P}
$$
(de-Broglie of matter waves)
and
$$
\lambda_2=\frac{h c}{E}
$$
From the Eqs. (ii) and (iii), we get
$$
\begin{aligned}
\frac{\lambda_1}{\lambda_2} & =\frac{h / P}{h c / E}=\frac{E}{P_C} \\
& =\frac{E}{\sqrt{2 m E} \cdot C}=\frac{\sqrt{E}}{\sqrt{2 m} \cdot C}
\end{aligned}
$$
$\begin{array}{ll}\text { So, } & \frac{\lambda_1}{\lambda_2}=\frac{\sqrt{E}}{\sqrt{2 m} \cdot c} \\ & \frac{\lambda_1}{\lambda_2} \propto \sqrt{E} \\ \Rightarrow & \lambda_1: \lambda_2=E^{1 / 2}\end{array}$

Topic: Physics, Dual Nature of Matter, JEE Main
The energy of a system as a function of time t is given as Et=A2-Î±t , where a=0.2 s-1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t=5 s is

Options:
A 4
B 8
C 2
D 12

The correct answer is:
4

Solution:
Energy E=A2e-Î±t
For small % errors, we can, do differentiation
dE=2AdAe-Î±t+A2-Î±e-Î±tdt
Fractional error =dEE=2Ae-Î±tdA+-Î±A2e-Î±tdtA2e-Î±t=2dAA+-Î±dttt
% error =21.25%+0.2Ã1.5%Ã5
=4% (errors always add up)
Alternate solution:
E=A2e-Î±t
Taking natural logarithm on both sides,
lnE=lnA2+-Î±t
Differentiating
dEE=2dAA+-Î±dt
For small fractional erros, errors always add up
dEE=2dAA+Î±dttÃt
=21.25%+0.21.5%5
=4%

Topic: Physics, Units and Dimensions, JEE Main
The energy of activation for a reaction is 100 kJ mol-1 . Presence of a catalyst lowers the activation energy by 75%. What will be effect on rate of reaction at 20oC , other things being equal?

Options:
A increases by a factor of
2.34Ã1013
B increases by a factor of
2.34× 10 10
C increases by a factor of 10 15
D no effect

The correct answer is:
increases by a factor of
2.34×1013

Solution:
The Arrhenius equation is
k=Ae-Ea/RT
In absence of catalyst, k1=Ae-100/RT
In presence of catalyst, k2=Ae-25/RT
So k2k1=e-75/RT or 2.303 logk2k1=75RT
or 2.303 logk2k1=758.314Ã10-3Ã293
or logk2k1=758.314Ã10-3Ã293Ã2.303
or k2k1=2.34Ã1013
As the things being equal in presence or absence of a catalyst,
k 2 = rate in presence of catalyst
k 1 = rate in absence of catalyst
i.e., r2r1=k2k1=2.34Ã1013

Topic: Chemistry, Chemical Kinetics, JEE Main
The energy of an electromagnetic wave contained in a small volume oscillates with

Options:
A Double the frequency of the wave
B The frequency of the wave
C Half the frequency of the wave
D Zero frequency

The correct answer is:
Double the frequency of the wave

Solution:
The formula to calculate the energy confined within the electric component of an electromagnetic wave can be written as

uE=12ε0E2dV ...1

Let's assume that the equation of the electric component is given by

E=E0 sinωt+ϕ ...2

Substitute the expression for the electric field component from equation (2) into equation (1) and simplify to obtain the expression for the energy confined within the small volume dV.

uE=12ε0E0 sinωt+ϕ2dV= 12ε0E02sin2ωt+ϕdV=14ε0E021-cos2ωt+2ϕ ...3

From equation (3), it can be concluded that the frequency of oscillation of the energy confined within a small volume is double the frequency of the wave.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The energy of an electron in Bohr's orbit of hydrogen atom is −13.6 eV . The total electronic energy of a hypothetical He atom in which there are no electron-electron repulsions or interactions is

Options:
A 27.2 eV
B -27.2 eV
C -108.8 eV
D 108.eV

The correct answer is:
-108.8 eV

Solution:
Electronic energy per electron =-13.6×4 eV
Total Electronic energy =-13.6 × 4 × 2=-108.8 eV

Topic: Chemistry, Structure of Atom, JEE Main
The energy of an electron in first Bohr's orbit of H atom is −13.6 eV. The energy value of electron in the first excited state of Li 2+ is :

Options:
A 27.2 eV
B â30.6âeV
C 30.6 eV
D −27.2 eV

The correct answer is:
−30.6 eV

Solution:
For H atom & H like species
The electronic energy in the nth orbit is
E n = - 13.6 z 2 n 2 eV
E n Li 2 + = - 13.6 Ã 9 n 2 eV = - 122.4 eV n 2
If n = 2
E 2 Li 2 + = - 122.4 2 2 = - 30.6 eV

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (09 Apr Online)
The energy of an electron in n th orbit of hydrogen atom is - 13.6 n⁡ 2 eV.⁡ Energy required to excite the electron from the first orbit to the third orbit is

Options:
A 10.2  J
B 12.09  J
C 12.09  eV
D 13.6  eV

The correct answer is:
12.09  eV

Solution:
Calculate difference in energies of 3rd orbit and 1st orbit.
Use, - 13.6 nâ¡ 2
E=E3-E1=-13.632â-13.612
E=-1.51+13.6=12.09 eV

Topic: Physics, Atomic Physics, JEE Main
The energy of an electron in the first Bohr orbit of hydrogen atom is -2.18×10-18 J Its energy in the third Bohr orbit is_______.

Options:
A 19th of the value
B 127 of this value
C

Three times of this value

D

One third of this value

The correct answer is:

Three times of this value

Solution:
The energy of an electron in the nth Bohr orbit of a hydrogen atom can be calculated using the formula:

En=-2.18×10-18Z2n2Where n is the principal quantum number of the Bohr orbit.Enα1n2Enα19

Therefore, the energy of the electron in the third Bohr orbit of a hydrogen atom is Three times of this value.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The energy of an electron in the first orbit of H - atom is -13.6 eV. The possible energy values of the excited state for electrons in Bohr orbits of Li2+ ions is/are-

Options:
A â3.0 eV
B â30.6Â eV
C â13.6Â eV
D Both B & C are correct

The correct answer is:
Both B & C are correct

Solution:
En=â13.6 Z2n2 for Li2+, En=â 13.6 Ã 9n2
= -122.4n2 eV
If n=1, En= â 122.4 eV
If n=2, En= â 30.6 eV
If n=3, En= â 13.6 eV

Topic: Chemistry, Structure of Atom, JEE Main
The energy of electron in the nth orbit of hydrogen atom is expressed as $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$.

The shortest and longest wavelength of Lyman series will be

Options:
A $910 Ã, 1213 Ã$
B $5463 Ã, 7858 Ã$
C $1315 Ã, 1530 Ã$
D None of these

The correct answer is:
$910 Ã, 1213 Ã$

Solution:
$\frac{1}{\lambda_{\max }}=\mathrm{R}\left[\frac{1}{(\mathrm{l})^{2}}-\frac{1}{(2)^{2}}\right]$

$\Rightarrow \lambda_{\mathrm{max}}=\frac{4}{3 \mathrm{R}} \approx 1213 Ã…$

and

$\frac{1}{\lambda_{\min }}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{\infty}\right] \Rightarrow \lambda_{\min }=\frac{1}{\mathrm{R}} \approx 910 Ã…$

Topic: Physics, Atomic Physics, JEE Main
The energy of He+ ion in its first state is, (The ground state energy for the Hydrogen atom-13.6 eV):

Options:
A -27.2 eV
B -3.4 eV
C -13.6 eV
D -54.4 eV

The correct answer is:
-13.6 eV

Solution:
Energy of Hydrogen like atoms is, E=-13.6Z2n2eV.

Since Helium is in the first state, Z=2, n=2.

The energy of He+ ion is,

E=-13.6×2222=-13.6 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The energy of one mole of photons of radiation of frequency 2×1012 Hz in Jmol-1 is _____ .
(Nearest integer)
(Given: h=6.626×10-34Js NA=6.022×1023 mol-1)

Options:

The correct answer is:
798

Solution:
For one photon, from Planck's equation, E=hv

For one mole photon,

⇒E=6·023×1023×6·626×10-34×2×1012

=798·16 J

≈798 J

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The energy of one mole of photons of radiation of wavelength 300 nm is (Given : h=6.63×10-34 J s,NA=6.02×1023 mol-1c=3×108 m s-1)

Options:
A 235 kJ mol-1
B 325 kJ mol-1
C 399 kJ mol-1
D 435 kJ mol-1

The correct answer is:
399 kJ mol-1

Solution:
λ=300×10−9m

Energy of single photon =hcλ

Energy of NA photons =6.02×1023×6.63×10−34×3×108300×10−9

= 3.99×105 J/mole=399 KJ/mole

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The energy of photon of wavelength \( \lambda \) is

Options:
A \( \frac{c \lambda}{h} \)
B \( \frac{h \lambda}{c} \)
C \( \frac{h c}{\lambda} \)
D \( \frac{c}{h \lambda} \)

The correct answer is:
\( \frac{h c}{\lambda} \)

Solution:
The energy of a photon is given by,

E=hν ...1;

where h is Planck constant and

ν is the frequency of light.

The wavelength of the photon is given by,

λ=cν ...2;

where c is speed of light.

From equation 1 and 2,

E=hcλ

The energy of a photon E=hcλ.

Topic: Physics, Dual Nature of Matter, JEE Main
The energy of second Bohr orbit of the hydrogen atom is $-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$; hence the energy of fourth Bohr orbit would be

Options:
A $-41 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-164 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\mathrm{Z}=1$ for hydrogen ; $n=2$
$$
\begin{aligned}
& \mathrm{E}_n=-k\left(\frac{\mathrm{Z}}{n}\right)^2 \\
& \mathrm{E}_2=\frac{-k \times 1}{4} \\
& \mathrm{E}_2=-328 \mathrm{KJ} \mathrm{mol}^{-1} ; k=4 \times 328 \\
& \mathrm{E}_4=\frac{-k \times 1}{16} \\
& \mathrm{E}_4=-4 \times 328 \times \frac{1}{16} \\
& \mathrm{E}_4=-82 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$$
Related Theory
The total mechanical energy of an electron in a Bohr orbit is the sum of its kinetic and potential energies. The lowest energy is called the ground state.
Caution
It is given that the energy of the electron in the first Bohr $\mathrm{H}$-atom is $-13.6 \mathrm{eV}$

Topic: Chemistry, Structure of Atom, JEE Main
The energy of second Bohr orbit of the hydrogen atom is â 328 kJ mol â1 The energy of fourth Bohr orbit would be:

Options:
A − 41 kJ mol −1
B − 82 kJ mol −1
C − 164 kJ mol −1
D − 1312 kJ mol −1

The correct answer is:
− 82 kJ mol −1

Solution:
We know that E n  =  E 1  ×  Z n 2
Given E 2  =  E 1 2 2  = − 328 kJ mol −1
E 4  =  E 2 2 2  =  −328 4   kJ mol −1  = − 82 kJ mol −1

Topic: Chemistry, Structure of Atom, JEE Main
The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong?

Options:
A Infra-red rays
B Ultraviolet rays
C γ -rays
D X-rays

The correct answer is:
X-rays

Solution:
Wavelength of the ray
λ=hcE=1240015000=0.826 Å
X-ray wavelengths are in the range 0.5 – 2.5 Å, so it is an X-ray.

Topic: Physics, Electromagnetic Waves, JEE Main
The energy of X-ray photon is \( 2200 \mathrm{eV} \). Its frequency would be

Options:
A \( 5.3 \times 10^{16} \mathrm{~Hz} \)
B \( 5.3 \times 10^{17} \mathrm{~Hz} \)
C \( 5 \times 10^{17} \mathrm{~Hz} \)
D \( 5 \times 10^{16} \mathrm{~Hz} \)

The correct answer is:
\( 5.3 \times 10^{17} \mathrm{~Hz} \)

Solution:
According to Planck's hypothesis, the energy of a photon E is proportional to the frequency ν of photon where proportionality constant is Planck's constant h. So,E=hν⇒ν=Eh =2200×1.6×10-196.63×10-34 =5.3×1017Hz

Topic: Physics, Electromagnetic Waves, JEE Main
The energy released by the fission of one uranium atom is $200 \mathrm{MeV}$. The number of fissions per second required to produce $3.2 \mathrm{~W}$ of power is (Take $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ )

Options:
A $10^7$
B $10^{10}$
C $10^{15}$
D $10^{11}$

The correct answer is:
$10^{11}$

Solution:
Hints: $u=200 \mathrm{MeV}=200 \times 10^6 \mathrm{eV}=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}$
$\mathrm{E}=3.2 \mathrm{~J}$
No of fissions $=\frac{3.2}{2 \times 1.6 \times 10^{-11}}=10^{11}$

Topic: Physics, Nuclear Physics, JEE Main
The energy released during a nuclear reaction is called ______ energy

Options:
A Electrostatic Energy
B Fission Energy
C Potential Energy
D Binding Energy

The correct answer is:
Binding Energy

Solution:
The energy released during a nuclear reaction is called binding energy.
Binding energy is the minimum energy required to disassemble a system of particles into separate parts.

Topic: Physics, Nuclear Physics, JEE Main
The energy released during conversion of million atoms of iodine in gaseous state to iodide ions in gaseous state is 4.9×10-13J. What is the electron gain enthalpy in eV/atom.
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
3

Solution:
â´106 atoms of Ig=4.9Ã10-13J
∴6.023× 10 23 atoms  of I g = 4.9× 10 −13 ×6.023× 10 23 J 10 6
E=295.127 kJ/mole
Also E=295.127Ã1031.60Ã10-19Ã6.023Ã1023â3 eV

Topic: Chemistry, Thermodynamics (C), JEE Main
The energy releasing process in which the substrate is oxidised without an external electron acceptor is called

Options:
A Glycolysis
B Fermentation
C Photorespiration
D Aerobic respiration

The correct answer is:
Fermentation

Solution:
Fermentation consumes $\mathrm{NADH}_2$ and $\mathrm{e}^{-}$acceptor is released as $\mathrm{NAD}^{+}$not used. While glycolysis uses external $\mathrm{e}^{-}$acceptor as $\mathrm{NAD}^{+}$.

Topic: Biology, Respiration in Plants, NEET, NEET 2008 (Mains)
The energy required to break one mole of Cl-Cl bonds in Cl2 is 242 kJ mol-1 . The longest wavelength of light capable of breaking a single Cl-Cl bond is

( C=3×108ms-1 and NA=6.02×1023mol-1 )

Options:
A 594 nm
B 640 nm
C 700 nm
D 494 nm

The correct answer is:
494 nm

Solution:
Energy required to break one mole of Cl-Cl bonds in Cl2
=242Ã1036.023Ã1023=hcÎ»=6.626Ã10-34Ã3Ã108Î»
So Î»=6.626Ã10-34Ã3Ã108Ã6.023Ã1023242Ã103
=0.4947Ã10-6 m
=494.7 nm

Topic: Chemistry, Structure of Atom, JEE Main
The energy required to break one mole of $\mathrm{Cl}-\mathrm{Cl}$ bonds in $\mathrm{Cl}_2$ is $242 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The longest wavelength of light capable of breaking a single $\mathrm{Cl}-\mathrm{Cl}$ bond is $\left(\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\right.$ and $\left.\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)$

Options:
A
$594 \mathrm{~nm}$

B
$640 \mathrm{~nm}$

C
$700 \mathrm{~nm}$

D
$494 \mathrm{~nm}$

The correct answer is:
$494 \mathrm{~nm}$

Solution:
Energy required for $1 \mathrm{Cl}_2$ molecule $=\frac{242 \times 10^3}{\mathrm{~N}_{\mathrm{A}}}$ Joules.
This energy is contained in photon of wavelength ' $\lambda$ '.
$$
\begin{aligned}
& \frac{\mathrm{hc}}{\lambda}=\mathrm{E} \Rightarrow \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{\lambda}=\frac{242 \times 10^3}{6.022 \times 10^{23}} \\
& \lambda=4947 Å \approx 494 \mathrm{~nm}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2010
The energy required to break one mole of hydrogen-hydrogen bonds in $\mathrm{H}_{2}$ is $436 \mathrm{kJ}$. What is the longest wavelength of light required to break a single hydrogenhydrogen bond?

Options:
A $68.5 \mathrm{nm}$
B $137 \mathrm{nm}$
C $274 \mathrm{nm}$
D $548 \mathrm{nm}$

The correct answer is:
$274 \mathrm{nm}$

Solution:
Given, energy of one mole of $\mathrm{H}_{2}$ bonds $=436 \mathrm{k} \mathrm{J}$
$$
\begin{array}{l}
\because E=\frac{n \cdot h c}{\lambda}\\
\therefore \quad \lambda=\frac{n.h c}{E} \\
\lambda=\frac{6.634 \times 10^{-} 34 \mathrm{s} \times 3 \times 10^{3} \mathrm{ms}^{-1} \times 6.023 \times 10^{23}}{436 \times 10^{3} \mathrm{J}} \mathrm{m} \\
=0.2736 \times 10^{-34+8+23-3} \\
=0.2736 \times 10^{-6} \mathrm{m} \\
=273.6 \mathrm{nm}=274 \mathrm{nm}
\end{array}
$$

Topic: Chemistry, Structure of Atom, JEE Main