datavorous/entrance-exam-dataset huggingface dataset:
An electron moving with speed v and a photon moving with speed c, have the same D-Broglie wavelength. The ratio of the kinetic energy of the electron to that of a photon is:

Options:
A 3cv
B v3c
C v2c
D 2cv

The correct answer is:
v2c

Solution:
λe=λPh

hpe=hpph

2mke=Ephc

2mke=Eph2c2

keEph=Ephc212 m

=pphc12 m

=pec12 m

=mvc12 m

=v2c

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Jul Shift 2)
An electron of a hydrogen like atom, having Z = 4, jumps from 4th energy state to 2nd energy state, The energy released in this process, will be: (Given Rch = 13.6 eV)

Where R = Rydberg

constant c = Speed of light in vacuum

h = Planck's constant

Options:
A 13.6 eV
B 10.5 eV
C 3.4 eV
D 40.8 eV

The correct answer is:
40.8 eV

Solution:
According to Bohr's atomic theory, whenever an electron jumps from one orbit to another, there is always an emission or absorption energy in terms of emission or absorption of photons.

The formula to calculate the energy ∆E of the emitted or absorbed photon when an electron jumps from a state with quantum number n1 to another with quantum number n2 is given by

∆E= -RchZ21ni2-1nf2....................(1)

where, R is Rydberg's constant, h is Planck's constant, c is the speed of light and Z is the atomic number.

Substitute the values of the given parameters into equation (1) to calculate the required energy.

∆E=-13.6×42142-122 eV= -13.6×16×116-14 eV= 40.8 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (01 Feb Shift 2)
An electron of a stationary hydrogen atom (mass = m) jumps from the fifth energy level to the ground level. As a result, the atom emits a photon and recoils with some velocity. The velocity that the atom acquires as a result of photon emission is [h = Planck's constant, R = Rydberg's constant]

Options:
A 24hR25m
B 25hR24m
C 24m25hR
D 25m24hR

The correct answer is:
24hR25m

Solution:
Here, E5âE1=hcÎ»
â(Rhc52âRhc)=hcÎ»
(where R=Rydberg's constant, h=Planck's constant and c=speed of light in vacuum)
â2425R=1Î»
But P=hÎ»
âv=hmÎ»=24Rh25m

Topic: Physics, Atomic Physics, JEE Main
An electron of an atom transits from $n_{1}$ to $n_{2}$. In which of the following maximum frequency of photon will be emitted?

Options:
A $n_{1}=1$ to $n_{2}=2$
B $n_{1}=2$ to $n_{2}=1$
C $n_{1}=2$ to $n_{2}=6$
D $n_{1}=6$ to $n_{2}=2$

The correct answer is:
$n_{1}=2$ to $n_{2}=1$

Solution:
For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible.
In option (b), $n_{1}=2, n_{2}=1$
$\therefore$ Energy of emitted photon is given as
$$
\begin{aligned}
E^{\prime} &=-13.6 Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\Rightarrow \quad h v^{\prime} &=-13.6 Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right) \\
v^{\prime} &=\frac{13.6 Z^{2}}{h} \times \frac{3}{4}
\end{aligned}
$$
In option (c), $n_{1}=2, n_{2}=6$, hence energy of emitted photon is given as
$$
\begin{aligned}
& E^{\prime \prime}=-13.6 Z^{2}\left(\frac{1}{6^{2}}-\frac{1}{2^{2}}\right) \\
\Rightarrow & h v^{\prime \prime}=13.6 Z^{2} \times \frac{8}{36} \\
\Rightarrow & \quad v^{\prime \prime}=\frac{13.6 Z^{2}}{h} \times \frac{8}{36}
\end{aligned}
$$
Hence, from Eqs. (i) and (ii), we get
$$
v^{\prime}>v^{\prime \prime}
$$

Topic: Physics, Atomic Physics, JEE Main
An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{\mathbf{i}}$ is subjected to all electric field $E_0 \hat{\mathbf{j}}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $\left.=\lambda_0\right)$

Options:
A $\lambda_0$
B $\lambda_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$
C $\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$
D $\frac{\lambda_0}{\left(1+\frac{e^2 E_0^2 t^2}{m v_0^2}\right)}$

The correct answer is:
$\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$

Solution:
$\because$ de-Broglie relation of a charged particles,
$$
\lambda=\frac{h}{m v}
$$
Velocity of charged particle at time $t$,
$$
\mathbf{v}=v_0 \hat{\mathbf{i}}+\frac{e E_0}{m} t \hat{\mathbf{j}} \text { or }|v|=\sqrt{v_0^2+\left(\frac{e E_0}{m} t\right)^2}
$$
Hence, $\lambda=\frac{h}{m \sqrt{v_0^2+\left(\frac{e E_0}{m} t\right)^2}}$
or
$$
\begin{gathered}
\lambda=\frac{h}{m v_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \text { or } \lambda=\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \\
\left(\because \lambda_0=\frac{h}{m v_0}\right)
\end{gathered}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
An electron of charge e is moving in a circular orbit of radius r around a nucleus, at a frequency υ. The magnetic moment associated with the orbital motion of the electron is

Options:
A 2πυer2
B πυr22
C πeυr2
D πυr24

The correct answer is:
πeυr2

Solution:
The equivalent current due to the motion of the electron is

i=eυ

So, the magnetic moment will be

M=i×πr2=πeυr2

Topic: Physics, Magnetic Properties of Matter, JEE Main
An electron of hydrogen atom on an excited state is having energy En=-0.85 eV. The maximum number of allowed transitions to lower energy level is _______.

Options:

The correct answer is:
6

Solution:
Energy of nth orbit is given by, En=-13.6n2

⇒-13.6n2=-0.85⇒n=4

No of transition,

=n(n-1)2=4(4-1)2=6

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
An electron of mass 9.0Ã10-31 kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3Ã106 m/s. If a proton of mass 1.8Ã10-27 kg has to move in a circle of same radius and in the same magnetic field, then its speed must be

Options:
A 1.5Ã103 m/s
B 3×106 m/s
C 6×104 m/s
D 2×108 m/s

The correct answer is:
1.5Ã103 m/s

Solution:
Magnetic force will provide the necessary centripetal force.
Bqv=mv2r
Bqr=mv
For electron and proton, the magnetic field B, charge q and radius r are same.
So, mv = constant
i.e. meve=mpvp
vp=mempve=9×10-311.8×10-27×3×106
vp=1.5×103 m/s

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)

Options:
A 1c2mE12
B 1cE2m12
C (E2 m)12
D c(2mE)12

The correct answer is:
1cE2m12

Solution:
λ1=h2mE

λ2=hcE

λ1λ2=1cE2m1/2

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (17 Mar Shift 1)
An electron (of mass m ) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ( c= speed of light in vacuum)

Options:
A 1c2Em1/2
B c(2mE)1/2
C 1cE2m1/2
D E2m1/2

The correct answer is:
1cE2m1/2

Solution:
λd for electron =h2mE
λ for photon =hCE
Ratio =h2mEEhC=1CE2m

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (07 Jan Shift 2)
An electron of mass ' $m$ ' and charge ' $q$ ' is accelerated from rest in a uniform electric field of strength ' $E$ '. The velocity acquired by the electron when it travels a distance ' $L$ ' is

Options:
A $\sqrt{\frac{2 q E}{m L}}$
B $\sqrt{\frac{2 E m}{q L}}$
C $\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$
D $\sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}$

The correct answer is:
$\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$

Solution:
Force on the electron $\mathrm{F}=\mathrm{qE}$
Work done by the force $\mathrm{W}=\mathrm{qEL}$
Work done is equal to gain in kinetic energy
$$
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{qEL} \\
& \therefore \mathrm{v}=\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
An electron of mass m and electric charge e is accelerated by a potential difference of V and then it enters in a magnetic field B perpendicular to the magnetic field lines. Radius of the circular path formed by electron is

Options:
A 2 eV/m
B 2 Vm/eB2
C 2 Vm/eB
D 2 Vm/e2B

The correct answer is:
2 Vm/eB2

Solution:
Bev=mv2râââorâââr=mvBe as mv=2mT (T = KE). So r=2mTBe

As the electron has been accelerated from rest through a potential difference of V volt, then T=eV

â´ r=2mVeB2e2=2mVB2e

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron of mass m and magnitude of charge e at rest, gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at a time t is (ignore relativistic effects)

Options:
A dλdt=-heEt
B dλdt=-2heEt
C dλdt=-2heEt2
D dλdt=-heEt2

The correct answer is:
dλdt=-heEt2

Solution:
The acceleration of the electron is

a=eEm

The velocity of the electron as a function of time is

v=at=eEmt

The de-Broglie wavelength is

λ=hmv

λ=hmeEmt

λ=heEt
dλdt=-heEt2

Topic: Physics, Dual Nature of Matter, JEE Main
An electron of mass me and a proton of mass mp=1836 me are moving with the same speed. The ratio of their de Broglie wavelength λelectronλproton will be :

Options:
A 1836
B 11836
C 918
D 1

The correct answer is:
1836

Solution:
Given mass of electron =me
Mass of proton =mp
∴ given mp=1836 me
From de-Broglie wavelength
λ=hp=hmv
λeλp=mpme
=1836 meme
λeλp=1836

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Feb Shift 2)
An electron of mass me and a proton of mass mP are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

Options:
A mPme
B 1
C mpme
D memp

The correct answer is:
mpme

Solution:
KE=eΔV
λe=h2 me(eΔV)
λp=h2 mp(eΔV)

⇒ λeλp=mpme

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (22 Jul Shift 1)
An electron of mass m has de-Broglie wavelength Î» when accelerated through potential difference V. When proton of mass M, is accelerated through potential difference 9V, the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)

Options:
A λ3Mm
B λ3.Mm
C λ3mM
D λ3.mM

The correct answer is:
λ3mM

Solution:
When electron or any charged particle is accelerated through potential difference V, then kinetic energy gained is given by E=eV ...... (i)
E=12mv02=p22m=h22m.λ2 ...... (ii)
∴ eV=h22m.λ2 ⇒λ=h2meV ....... (iii)
When proton of mas M is accelerated through a potential difference of 9V, then the de - Broglie wavelength obtained is
λ′=h2M e9V=h3×2MeV×mm
∴ λ′=λ3.mM

Topic: Physics, Dual Nature of Matter, JEE Main
An electron of mass \( m_{e} \) and a proton of mass \( m_{p}=1836 m_{e} \) are moving with the same speed. The ratio of their de Broglie wavelength \( \frac{\lambda_{\text {electron }}}{\lambda_{\text {proton }}} \) will be :

Options:
A \( 1836 \)
B \( \frac{1}{1836} \)
C \( 918 \)
D \( 1 \)

The correct answer is:
\( 1836 \)

Solution:
Given mass of electron =me
Mass of proton =mp
∴ given mp=1836 me
From de-Broglie wavelength
λ=hp=hmv
λeλp=mpme
=1836 meme
λeλp=1836

Topic: Physics, Dual Nature of Matter, JEE Main
An electron practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavelength =λ1Å . It then get retorted through 19 volts and then has a wavelength λ2Å . It further retardation through 3.2 volts changes the wavelength to λ3 . What is the value of λ3-λ2λ1 ?

Options:
A 2041
B 1063
C 2063
D 1041

The correct answer is:
2063

Solution:
λ=h2meV
λ1=h2me100=h102me=k10
k=h2me
λ2=h2me81=k9
λ3=h2me49=k7
λ3-λ2λ1=k7-k9k10=2k63k10=2063

Topic: Chemistry, Structure of Atom, JEE Main
An electron revolves in a circle of radius $0.4 Ã…$ with a speed of $10^6 \mathrm{~m} / \mathrm{s}$ in a hydrogen atom. The magnetic field produced at the centre of the orbit due to the motion of the electron (in Tesla) is : $\left[\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right.$, charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right]$

Options:
A $0.1$
B $1.0$
C $10$
D $100$

The correct answer is:
$10$

Solution:
$$
r=0.4 Ã…,=0.4 \times 10^{-10} \mathrm{~m}, v=10^6 \mathrm{~m} / \mathrm{s}
$$
The moving electron will constitute an electric current
Electric current $i=\frac{q}{t}$
$$
\begin{aligned}
i & =\frac{q}{\frac{2 \pi r}{v}}=\frac{q v}{2 \pi r} \\
i & =\frac{1.6 \times 10^{-19} \times 10^6}{2 \pi \times 0.4 \times 10^{-10}} \\
& =\frac{1.6 \times 10^{-3}}{0.8 \pi}
\end{aligned}
$$
Magnetic produced at the centre
$$
\begin{aligned}
B & =\frac{\mu_0 i}{2 r} \\
B & =\frac{4 \pi \times 10^{-7}}{2 \times 0.4 \times 10^{-10}} \times \frac{1.6 \times 10^{-3}}{0.8 \pi} \\
& =\frac{4 \pi \times 1.6}{0.8 \times 0.8 \pi}=\frac{6.4}{0.64} \\
& =10 \text { tesla }
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron revolving in nth Bohr orbit has magnetic moment μn. If μn ∝ nx, the value of x is:

Options:
A 2
B 1
C 3
D 0

The correct answer is:
1

Solution:
Magnetic moment of electron is given by, μn=ehn4πme.

Clearly, μn∝ n1. Therefore, x=1.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :

Options:
A $\mathrm{E}=\mathrm{U}$
B $2 \mathrm{E}=\mathrm{U}$
C $2 \mathrm{E}=3 \mathrm{U}$
D $\mathrm{E}=2 \mathrm{U}$

The correct answer is:
$2 \mathrm{E}=\mathrm{U}$

Solution:
$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{PE}=-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}}-\frac{\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{\mathrm{r}}=\frac{-\mathrm{K}(\mathrm{Ze})(\mathrm{e})}{2 \mathrm{r}} \\ & \mathrm{TE}=\frac{\mathrm{PE}}{2} \\ & 2 \mathrm{TE}=\mathrm{PE}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of 3 $\mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$, so that electron moves along the same path, is _______ $\mathrm{NC}^{-1}$.
$\left(\right.$ Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$

Options:

The correct answer is:
4

Solution:
For the given condition of moving undeflected, net force should be zero.
$\begin{aligned} \mathrm{qE} & =\mathrm{qVB} \\ \mathrm{E} & =\mathrm{VB} \\ & =\sqrt{\frac{2 \times \mathrm{KE}}{\mathrm{m}}} \times \mathrm{B} \\ & =\sqrt{\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \times 3 \times 10^{-6} \\ & =4 \mathrm{~N} / \mathrm{C}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (08 Apr Shift 1)
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle α with the plates. It leaves the plates at angle β with kinetic energy K2. Then the ratio of kinetic energies K1:K2 will be :

Options:
A sin2βcos2α
B cosβsinα
C cosβcosα
D cos2βcos2α

The correct answer is:
cos2βcos2α

Solution:
velocity along the plate will not change.

∴ v1cosα=v2cosβ

K1 K2⇒v12v22=cos2βcos2α

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (25 Feb Shift 2)
An electron with speed $v$ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and Pe and that of photon are Eph and Pph respectively. Which of the following is correct?

Options:
A EeEph=2cv
B EeEph=v2c
C pepph=2cv
D pepph=v2c

The correct answer is:
EeEph=v2c

Solution:
We have, λp=hpp and λe=hpe

Given here, λe=λp, then pp=pe

Ratio of their energy is EphEe=hcλPpe22m=2mcpe=2mcmv=2cv

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (26 Jun Shift 1)
An element A of group 1 shows similarity to an element B belonging to group 2. If A has maximum hydration enthalpy in group 1 then B is

Options:
A Mg
B Be
C Ca
D Sr

The correct answer is:
Mg

Solution:
Hydration enthalpy is inversely related to the size of the ion.

Li+ has maximum hydration enthalpy in group 1 due to its small size.

Lithium shows diagonal relationship to magnesium.

The similarity between lithium and magnesium is particularly striking and arises because of their similar sizes : atomic radii, Li = 152 pm, Mg = 160 pm; ionic radii : Li+ = 76 pm, Mg2+= 72 pm.

The main points of similarity are

(I)Both lithium and magnesium are harder and lighter (ii) Lithium and magnesium react slowly with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating. Both form a nitride, Li3N and Mg3N2 , by direct combination with nitrogen. (iii) The oxides, Li2O and MgO do not combine with excess oxygen to give any superoxide. (iv) The carbonates of lithium and magnesium decompose easily on heating to form the oxides and CO2 .

Solid hydrogencarbonates are not formed by lithium and magnesium. (v) Both LiCl and MgCl2 are soluble in ethanol. (vi) Both LiCl and MgCl2 are deliquescent and crystallise from aqueous solution as hydrates, LiCl·2H2O and MgCl2 ·6H2O.

So 'B' is Mg.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (27 Jul Shift 2)
An element belonging to chalcogen group is

Options:
A carbon
B phosphorus
C chlorine
D sulphur

The correct answer is:
sulphur

Solution:
Group 16 elements are called chalcogens.
$\therefore$ Sulphur belongs to chalcogen family.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
An element crystallises in a face–centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is:

Options:
A a2
B a
C 2a
D a2

The correct answer is:
a2

Solution:
In FCC octahedral voids are present at the edge centers and body center.

Minimum distance between centers of two octahedral voids

=x=a22+a22
=a24+a24=a2

Topic: Chemistry, Solid State, JEE Main, JEE Main 2020 (05 Sep Shift 2)
An element crystallizes both in fcc and bcc lattice. If the density of the element in the two forms is the same, the ratio of unit cell length of fcc to that of bcc lattice is

Options:
A (2)1/3
B 121/3
C (4)1/3
D 141/3

The correct answer is:
(2)1/3

Solution:
Z=a3×d×NAM
fcc (4) =a13×d×NAM …(i)
bcc (2) =a23×d×NAM …(ii)
Dividing equation (i) by (ii)
2=a1a23
a1a2=21/3

Topic: Chemistry, Solid State, JEE Main
An element has a face-centered cubic fcc structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is

Options:
A a2
B 32a
C a
D 2a

The correct answer is:
a2

Solution:
8 symmetrical cube
Distance between two TV=a-a4+a4
=a2
Correct option is C .

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (12 Apr Shift 1)
An element having an atomic radius of $0.14 \mathrm{~nm}$ crystallizes in an $f_{c c}$ unit cell. What is the length of a side of the cell ?

Options:
A
$0.56 \mathrm{~nm}$

B
$0.24 \mathrm{~nm}$

C
$0.96 \mathrm{~nm}$

D
$0.4 \mathrm{~nm}$

The correct answer is:
$0.4 \mathrm{~nm}$

Solution:
For a fcc unit cell
$\begin{aligned}
& r=\frac{\sqrt{2} a}{4} \\
& a=\frac{4 r}{\sqrt{2}}=2 \sqrt{2} \times 0.14=0.39 \approx 0.4 \mathrm{~nm}.
\end{aligned}$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2013 (09 Apr Online)
An element $\Delta l=\Delta x \hat{i}$ is placed at the origin and carries a large current $I=10 \mathrm{~A}$. The magnetic field on the $y$-axis at a distance of $0.5 \mathrm{~m}$ from the elements $\Delta x$ of $1 \mathrm{~cm}$ length is:

Options:
A $4 \times 10^{-8} \mathrm{~T}$
B $10 \times 10^{-8} \mathrm{~T}$
C $8 \times 10^{-8} \mathrm{~T}$
D $12 \times 10^{-8} \mathrm{~T}$

The correct answer is:
$4 \times 10^{-8} \mathrm{~T}$

Solution:
$\begin{aligned} \overrightarrow{\mathrm{dB}} & =\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{(\overrightarrow{\mathrm{d} l} \times \overrightarrow{\mathrm{r}})}{\mathrm{r}^3}(\text { Tesla) } \\ & =\frac{10^{-7} \times 10 \times\left(\frac{1}{2} \times \frac{1}{100}\right)(+\hat{\mathrm{k}})}{\left(\frac{1}{2}\right)^3}=4 \times 10^{-8} \mathrm{~T}(+\hat{\mathrm{k}})\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (06 Apr Shift 1)
An element occurring in the $b c c$ structure has $12.08 \times 10^{23}$ unit cells. The total number of atoms of the element in these cells will be

Options:
A $24.16 \times 10^{23}$
B $36.18 \times 10^{23}$
C $6.04 \times 10^{23}$
D $12.08 \times 10^{23}$

The correct answer is:
$24.16 \times 10^{23}$

Solution:
There are two atoms in a $b c c$ unit cell.
So, number of atoms in $12.08 \times 10^{23}$ unit cells $=2 \times 12.08 \times 10^{23}=24.16 \times 10^{23}$ atom .

Topic: Chemistry, Solid State, JEE Main
An element of atomic number 29 belongs to

Options:
A s -block
B p -block
C d -block
D f-block

The correct answer is:
d -block

Solution:
$\mathrm{Cu}_{29}-[\mathrm{Ar}] 3 d^{10} 4 s^1$. Since the last electron enters in the \(d\) subshell, it is a \(d\) block element.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An element P (atomic weight = 80 g/mol) having fcc structure, calculate number of unit cells in 8 gm of P

Options:
A 0.4×NA
B 0.1×NA
C 4×NA
D None of these

The correct answer is:
None of these

Solution:
Number of atoms in fcc = 4
Total number of atoms =watomic wt.×NA
Total number of atoms =880×NA=NA10
Number of unit cell =NA10×14
=6.04×1022
=1.5×1022

Topic: Chemistry, Solid State, JEE Main
An element undergoes a reaction as shown :

X + 2 e - â X 2 - , energy released =30.87âeV/atom. If the energy released is used to dissociate 4 gms of H2 molecules, equally into H+ and H*, where H* is excited state of H atoms where the electron travels in orbit whose circumference equal to four times its de Broglie's wavelength. Determine the least moles of X that would be required:

Given: I.E. of H=13.6âeV/atom bond energy of H2=4.526âeV/molecule.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Total energy requirement per molecule of H 2 :
Binding energy =4.526 eV
I.Eâ=â13.60 âeV
According to de Broglie's explanation of Bohr's orbit-
2Ïr n = nÎ»
As per question, 2Ïr = 4Î»
Thus, electron is excited to the 4th Bohr orbit
Excitation energy =E4âE1=â0.85ââ13.60=12.75eV
Thus, total eenergy requirement per molecule =13.60+12.75+4.526â30.87eV
Thus, 1 molecule of H 2 will be dissociated by 1 atom of X
Therefore, 4 gH2â¡2 moles of H2 molecules will be dissociated by 2 moles of X.

Topic: Chemistry, Structure of Atom, JEE Main
An element which never has a positive oxidation number in any of its compounds

Options:
A Boron
B Oxygen
C Chlorine
D Fluorine

The correct answer is:
Fluorine

Solution:
Fluorine always shows - 1 oxidation state.

Topic: Chemistry, Redox Reactions, JEE Main
An element with atomic number 20 will be placed in which period of the periodic table

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2$ principal quantum no. is 4 so it belongs to $4^{\text {th }}$ period.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An element with atomic number 21 is a

Options:
A halogen
B representative element
C transition element
D alkali metal

The correct answer is:
transition element

Solution:
The electronic configuration of element with atomic number 21 is
$$
1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{6}, 4 \mathrm{~s}^{2}, 3 \mathrm{~d}^{1}
$$
Since this element contains partly filled orbital, so it is a -block element. -block elements are also known as transition elements.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An element with molar mass 2.7×10−2 kg mol−1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×103 kg m-3, the radius of the element is approximately ……….×10-12m (to the nearest integer)

Options:

The correct answer is:
143

Solution:
Density = Z×MNa×Volume

d=Z×276.02×1023×4.05×10-33

2.7=Z×276.02×1023×4.05×10-33

Z=4 ⇒fcc unit cell

For fcc

unit cell 4r=2a

r=1.414×4054

=143.1675 pm

=143.17 pm

Topic: Chemistry, Solid State, JEE Main, JEE Main 2020 (03 Sep Shift 1)