datavorous/entrance-exam-dataset huggingface dataset:
An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×106 m s–1 in an orbit of radius 0.52 Å. The magnetic field produced at the nucleus of the hydrogen atom is _______ T.

Options:

The correct answer is:
40

Solution:
The magnetic field due to the electron is given by

B=μ0qvsin90°4πr2

The given data is

v=6.76×106 m s-1r=0.52×10-10 m

Thus, the magnetic field is

B=10-7×1.6××10-19×6.76×1060.52×0.52×10-20=40 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (15 Apr Shift 1)
An electron in a hydrogen atom undergoes a transition from a higher energy level to a lower energy level. The incorrect statement of the following is

Options:
A Kinetic energy of the electron increases
B Velocity of the electron increases
C Angular momentum of the electron remains constant
D Wavelength of de-Broglie wave associated with the motion of electron decreases

The correct answer is:
Angular momentum of the electron remains constant

Solution:
As the radius of the orbit will change due to transition from a higher energy level to a lower energy level.
Therefore angular momentum $\mathrm{L}=m v r$ will not remain constant.

Topic: Physics, Atomic Physics, AP EAMCET, AP EAMCET 2016
An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of $-3.4 \mathrm{eV}$. The kinetic energy and the de-Broglie wavelength of the electron are respectively

Options:
A $+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$
B $-3.4 \mathrm{eV}, 1.99 \times 10^{-9} \mathrm{~m}$
C $2.8 \mathrm{eV}, 2.38 \times 10^{-10} \mathrm{~m}$
D $1.1 \mathrm{eV}, 1.28 \times 10^{-9} \mathrm{~m}$

The correct answer is:
$+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$

Solution:
$E_{n}=-3.4 \mathrm{eV}$
The kinetic energyis equal to the magnitude of total energy in this case. $\therefore \mathrm{K.E}=+3.4 \mathrm{eV}$
The de Broglie wavelength of electron
$\begin{aligned} \lambda=\frac{h}{\sqrt{2 m K}} &=\frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \mathrm{eV} \\ &=0.66 \times 10^{-9} \mathrm{~m} \end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
An electron in an atom jumps in such a way that its kinetic energy changes from x to x9. The change in its potential energy (magnitude) will be-

Options:
A x9
B 16x9
C 9x
D 9x16

The correct answer is:
16x9

Solution:
Change in kinetic energy is
xâ x 9 = 8x 9
For Bohr model of atom
The kinetic energy in a state is equal to half of the potential energy in magnitude.
â´Potential energy
=2 Kinetic energy
2Ã8x9 =16x9

Topic: Chemistry, Structure of Atom, JEE Main
An electron in an excited state of Li2+ ion has angular momentum 3h2Ï. The de Broglie wavelength of the electron in this state is [ a0 is the Bohr radius]

Options:
A λ=2πa0
B Î»=4Ïa0
C Î»=Ïa0
D Î»=3Ïa0

The correct answer is:
λ=2πa0

Solution:
From Bohr's law
mvr=nh2π=3h2π
⇒n=3
And momentum =mv=3h2πr
Now, radius of nth shell, r=n2za0
⇒r=323.a0 ∵ZLi=3
⇒r=3a0
From De Broglie law
wavelength=hMomentum
⇒λ=hmv=h3h2πr
⇒λ=2πr3=2π3×3a0
λ=2πa0

Topic: Physics, Dual Nature of Matter, JEE Main
An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function $2.75 \mathrm{eV}$. If the stopping potential of the photo-electron is $10 \mathrm{~V}$, the value of $n$ is

Options:
A 3
B 4
C 5
D 2

The correct answer is:
3

Solution:
$$
\begin{aligned}
E & =\mathrm{KE}_{\max }+W \\
& =e V_0+W \\
& =10+2.75 \\
E & =12.75 \mathrm{eV}
\end{aligned}
$$
Difference of 4 and 1 energy level is $12.75 \mathrm{eV}$. So, higher energy level is 4 to ground and excited state is $n=3$.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Mains)
An electron in the hydrogen atom jumps from the first excited state'to the ground state. What will be the percentage change in the speed of electron?

Options:
A $25 \%$
B $50 \%$
C $75 \%$
D $100 \%$

The correct answer is:
$50 \%$

Solution:
Velocity of electron in the $\mathrm{n}^{\text {th }}$ orbit is
$$
\begin{aligned}
& \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{nh}} \\
& \Rightarrow \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}}
\end{aligned}
$$
Taking the ratio,
$$
\frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \Rightarrow \mathrm{v}_2=\frac{\mathrm{v}_1}{2}
$$
Change in velocity,
$$
\begin{aligned}
\Delta \mathrm{v} & =\left|\mathrm{v}_2-\mathrm{v}_1\right| \\
& =\left|\frac{\mathrm{v}_1}{2}-\mathrm{v}_1\right|=0.5 \mathrm{v}_1
\end{aligned}
$$
Therefore, change in percentage is $50 \%$

Topic: Physics, Atomic Physics, JEE Main
An electron initially at rest is accelerated through a potential difference of \( 1 \mathrm{~V} \). The energy acquired by electron is

Options:
A \( 10^{-19} \mathrm{~J} \)
B \( 1.6 \times 10^{-19} \) erg
C \( 1.6 \times 10^{-19} \mathrm{~J} \)
D \( 1 \mathrm{~J} \)

The correct answer is:
\( 1.6 \times 10^{-19} \mathrm{~J} \)

Solution:
Work required to move one electron (e) through a potential difference of 1 V, gives the unit of energy abbreviate as eV.
That is , Energy= eV
Given, e=1.6×10-19 C, V=1 volt
∴ Energy=1.6×10-19×1=1.6×10-19 J

Topic: Physics, Dual Nature of Matter, JEE Main
An electron is accelerated through a potential difference of $45.5 \mathrm{~V}$. The velocity acquired by it is $\left(\right.$ in $\mathrm{ms}^{-1}$ )

Options:
A $10^{6}$
B zero
C $4 \times 10^{6}$
D $4 \times 10^{4}$

The correct answer is:
$4 \times 10^{6}$

Solution:
As we know,
Kinetic energy of electron = Energy acquired due to
$$
\begin{aligned}
\Rightarrow \quad \frac{1}{2} m_{e} v^{2}=e V_{0} & \text { potential }\left(V_{0}\right) \\
\Rightarrow \quad v &=\sqrt{\frac{2 e V_{0}}{m_{e}}} \\
&=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}} \\
&=\sqrt{16 \times 10^{12}}=4 \times 10^{6} \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.
(A) The electron will experience magnetic force along the axis of the solenoid.
(B) The electron will not experience magnetic force.
(C) The electron will continue to move along the axis of the solenoid.
(D) The electron will be accelerated along the axis of the solenoid.
(E) The electron will follow parabolic path-inside the solenoid.
Choose the correct answer from the option given below:

Options:
A B, C and D only
B A and D only
C B and C only
D B and E only

The correct answer is:
B and C only

Solution:
The Lorentz force is given by F→=q(v→×B→)=qvBsinθn^. The angle θ=0°. Hence the force experienced is zero.

Thus, the electron will not experience any magnetic force and the electron will continue to move along the axis.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (11 Apr Shift 2)
An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is E→=30j^sin1.5×107t-5×10-2x V m-1. where t in in seconds and x is im meters.The maximum magnetic force experienced by the electron will be: (given c=3×108 m s-1 and electron charge =1.6×10-19 Coloumbs

Options:
A 3.2×10-18 N
B 2.4×10-18 N
C 4.8×10-19 N
D 1.6×10-19 N

The correct answer is:
4.8×10-19 N

Solution:
In electromagnetic wave is E0B0=C
so maximum value of magnetic field is

B0=E0C

Fmax=qVBmax·sin90°

=qV0E0C N

1.6×10-19×0.1×3×108×303×108=4.8×10-19 N

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (05 Sep Shift 1)
An electron is moving along the positive x-axis. If uniform magnetic field is applied parallel to the negative z-axis, then

A.The electron will experience magnetic force along positive y-axis

B.The electron will experience magnetic force along negative y-axis

C.The electron will not experience any force in magnetic field

D.The electron will continue to move along the positive x-axis

E.The electron will move along circular path in magnetic field

Choose the correct answer from the option given below:

Options:
A

A and E only

B

C and D only

C

B and E only

D

B and D only

The correct answer is:

B and E only

Solution:
The Lorentz force is given by

F→=qv→×B→

=-ev→×B→

So, F^=i^×k^=-j^.

F→ is along -y axis

As the magnetic field is perpendicular to the velocity, motion would be circular.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (13 Apr Shift 2)
An electron is moving along +x direction with a velocity of 6×106ms-1. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:

Options:
A 3×10-4T, along +z direction
B 5×10-3T, along -z direction
C 5×10-3T, along +z direction
D 3×10-4T, along -zdirection

The correct answer is:
5×10-3T, along +z direction

Solution:
q∨B=qE

B=EV=300006×106=5×103T

Magnetic force F=q(V→×B→)

So Ans.5×103T ,along +Z direction

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (06 Sep Shift 1)
An electron is moving in a cyclotron at a speed of \( 3.2 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1} \) in a magnetic field \( 5 \times 10^{-4} \mathrm{~T} \) perpendicular to it. The frequency of this electron is: \( \left(q=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right) \)

Options:
A \( 1.4 \times 10^{5} \mathrm{~Hz} \)
B \( 1.4 \times 10^{7} \mathrm{~Hz} \)
C \( 1.4 \times 10^{6} \mathrm{~Hz} \)
D \( 1.4 \times 10^{9} \mathrm{~Hz} \)

The correct answer is:
\( 1.4 \times 10^{7} \mathrm{~Hz} \)

Solution:
v=3.2×107m s-1 ; B=5×10-4 T
The frequency of electron is
f=qB2πm=1.6×10-19×5×10-42×3.14×9.1×10-31
f=1.4×107Hz=14 MHz

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron is moving on a circular path around a stationary neutron under gravitation interaction. Mass of the neutron and electron are M and m respectively. If Bohrâs quantum condition holds here the minimum permissible de Broglie wavelength associated with the electron will be

Options:
A h22πGMm
B h22πGMm2
C h22πGM2m
D h24πGM2m

The correct answer is:
h22πGMm2

Solution:
mv2r=GMmr2

mvr=nh2π

λ=hmv

λmin=h22πGM2m

Topic: Physics, Dual Nature of Matter, JEE Main
An electron is moving with an initial velocity $v=v_0 \vec{i}$ and is in a magnetic field $\mathrm{B}=\mathrm{B}_0 \overrightarrow{\mathrm{J}}$. Then, it's de-Broglie wavelength

Options:
A
remains constant

B
increases with time

C
decreases with me

D
increases and decreases periodically

The correct answer is:
remains constant

Solution:
As given that, $v=v_0 \vec{i}$ and $B=B_0 \vec{j}$
Force on moving electron due to perpendicular magnetic field $B$ is, $F=-e ~(v \times B)$
$$
\begin{array}{ll}
\mathrm{F}=-\mathrm{e}\left[\mathrm{v}_0 \hat{\mathrm{i}} \times \mathrm{B}_0 \hat{\mathrm{i}}\right]=-\mathrm{ev}_0 \mathrm{~B}_0(\hat{\mathrm{i}} \times \hat{\mathrm{j}}) \\
\Rightarrow=-\mathrm{ev}_0 \mathrm{~B}_0 \hat{\mathrm{k}} & (\because \hat{\mathrm{k}}=\hat{\mathrm{i}} \times \hat{\mathrm{j}})
\end{array}
$$
So, the force is perpendicular to $\mathrm{v}$ and $\mathrm{B}$, both as the force is $\perp$ to the velocity so the magnitude of $\mathrm{v}$ will not change, so momentum is $(=\mathrm{mv})$ will remain same or constant in magnitude. Hence,
de-Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ remians constant.

Topic: Physics, Dual Nature of Matter
An electron is moving with an initial velocity \( \vec{V}=V_{0} \hat{i} \) and is in a uniform magnetic field
\( \vec{B}=B_{0} \hat{j} \). Then its de Broglie wavelength

Options:
A decreases with time
B remains constant
C increase and decreases periodically
D increases with time

The correct answer is:
remains constant

Solution:
(B)
de-Broglie wavelength \( \lambda=\frac{h}{m v} \)
Electron is entering perpendicular to magnetic field. It moves in a circular path. But its speed remains same. Therefore
de Broglie wavelength also remains constant.

Topic: Physics, Magnetic Effects of Current, KCET, KCET 2019
An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

Options:
A the electron will continue to move with uniform velocity along the axis of the solenoid.
B the electron will be accelerated along the axis.
C the electron path will be circular about the axis.
D the electron will experience a force at $45^{\circ}$ to the axis and execute a helical path.

The correct answer is:
the electron will continue to move with uniform velocity along the axis of the solenoid.

Solution:
Since $\vec{v} \| \vec{B}$ so force on electron due to magnetic field is zero. So it will move along axis with uniform velocity.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (04 Apr Shift 1)
An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?

Options:
A Kinetic energy potential energy and total energy decrease
B Kinetic energy and total energy decreased but potential energy increases
C Kinetic energy increases but potential energy and total energy decreases
D Kinetic energy decreases, potential energy increases but total energy remains the same.

The correct answer is:
Kinetic energy increases but potential energy and total energy decreases

Solution:
$$
\begin{aligned}
& \text { Potential energy }=-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Kinetic energy }=-\frac{1}{2}(\text { P.E. })=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Total energy } \left.\frac{1}{2} \text { (P.E. }\right)=-\frac{1}{8 \pi \varepsilon_0} \frac{\mathrm{ze}^2}{\mathrm{r}}
\end{aligned}
$$
As $r$ decreases K.E, increases but P.E. and T.E. decreases (since they are negative).

Topic: Physics, Atomic Physics, MHT CET, MHT CET 2021 (23 Sep Shift 1)
An electron makes a transition from an excited state to the ground state of a hydrogen - like atom. Then

Options:
A kinetic energy decreases, potential energy increases but total energy remains same
B kinetic energy and total energy decrease but potential energy increases
C its kinetic energy increases but potential energy and total energy decrease
D kinetic energy, potential energy and total energy decrease

The correct answer is:
its kinetic energy increases but potential energy and total energy decrease

Solution:
$\begin{aligned}
U=-K \frac{z e^{2}}{r} ; T . E=-\frac{k}{2} \frac{z e^{2}}{r} \\
K . E=\frac{k}{2} \frac{z e^{2}}{r} . \text { Here } r \text { decreases }
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
An electron makes a transition from an excited state to the ground state of a hydrogen - like atom. Then

Options:
A kinetic energy decreases, potential energy increases but total energy remains same
B kinetic energy and total energy decrease but potential energy increases
C its kinetic energy increases but potential energy and total energy decrease
D kinetic energy, potential energy and total energy decrease

The correct answer is:
its kinetic energy increases but potential energy and total energy decrease

Solution:
$\begin{aligned}
U=-K \frac{z e^{2}}{r} ; T . E=-\frac{k}{2} \frac{z e^{2}}{r} \\
K . E=\frac{k}{2} \frac{z e^{2}}{r} . \text { Here } r \text { decreases }
\end{aligned}$

Topic: Physics, Atomic Physics, VITEEE, VITEEE 2017
An electron makes a transition from an excited state to the ground state of a hydrogen-like atom. Then

Options:
A Kinetic energy decreases, potential energy increases but total energy remains same
B Kinetic energy and total energy decrease but potential energy increases
C Its kinetic energy increases but potential energy and total energy decrease
D Kinetic energy, potential energy and total energy decrease

The correct answer is:
Its kinetic energy increases but potential energy and total energy decrease

Solution:
U=−Kze2r; TE=−k2ze2r
KE=k2ze2r. Here r decreases

Topic: Physics, Atomic Physics, NEET
An electron makes a transition from an excited state to the ground state of a ydrogen like atom. out of the following statements which one is correct?

Options:
A K.E., P.E. and T.E. decreases.
B K.E. decreases, P.E. increases but total energy remains the same.
C E. increases but P.E. and T.E. decreases.
D K.E. and T.E. decreases but P.E. increases.

The correct answer is:
K.E. decreases, P.E. increases but total energy remains the same.

Solution:
$\begin{aligned}
\text { T.E. } &=-13.6 \frac{Z^{2}}{n^{2}} \mathrm{eV}, \text { P.E. }=2 \text { T.E. } \\
\text { and } k . E . &=13.6 \frac{Z^{2}}{n^{2}} \mathrm{eV}
\end{aligned}$
Hence if n decreases k.E increases but P.E. and T.E. decrease.

Topic: Physics, Atomic Physics, MHT CET, MHT CET 2020 (14 Oct Shift 2)
An electron (mass = 9.1 Ã 10-31 kg; charge = - 1.6 Ã 10-19 C) experiences no deflection if subjected to an electric field of 3.2 Ã 105 V m-1 and a magnetic field of 2.0 Ã 10-3 Wb m-2. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit radius

Options:
A 45 m

B 4.5 m
C 0.45 m

D 0.045 m

The correct answer is:
0.45 m

Solution:
F→e+F→m=0⇒Fe=Fm⇒eE=eVB
v=EB⇒3.2×1052×10-3=1.6×108 m s-1
r=mvqB=9.1×10-3×1.6×1081.6×10-19×2×10-3=0.45 m

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron (mass $=9.0 \times 10^{-31} \mathrm{~kg}$ and charge $=1.6 \times 10^{-19}$ coulomb) is moving in a circular orbit in a magnetic field of $1.0 \times 10^{-4}$ weber $/ \mathrm{m}^2$. Its period of revolution is

Options:
A $3.5 \times 10^{-7} \mathrm{sec}$
B $7.0 \times 10^{-7} \mathrm{sec}$
C $1.05 \times 10^{-6} \mathrm{sec}$
D $2.1 \times 10^{-6} \mathrm{sec}$

The correct answer is:
$3.5 \times 10^{-7} \mathrm{sec}$

Solution:
\(T=\frac{2 \pi m}{q B}=\frac{2 \times 3.14 \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 1 \times 10^{-4}}=3.5 \times 10^{-7} \mathrm{sec}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron (mass m) with an initial velocity v→=v0i^v0>0 is moving in an electric field E→=-E0i^E0>0 where E0 is constant. If at t=0, de-Broglie wavelength is λ0=hmv0, then its de-Broglie wavelength after time t is given by

Options:
A λ0
B λ01+eE0tmv0
C λ0t
D λ01+eE0tmv0

The correct answer is:
λ01+eE0tmv0

Solution:
Initial de-Broglie wavelength of electron,

λ0=hmv0 ....(i)

Force on electron in electric field,

F=-eE=-e-E0i^=eE0i^

Acceleration of electron, a=Fm=eE0i^m
Velocity of electron after time t,

v=v0i^+eE0i^mt=v0+eE0mti^=v01+eE0mv0ti^

de-Broglie wavelength associated with electron at time t is

λ=hmv

⇒λ=hmv01+eE0mv0t=λ01+eE0mv0t

Topic: Physics, Dual Nature of Matter, JEE Main
An electron (mass m) with an initial velocity v→=v0i^v0>0 is moving in an electric field E→=-E0i^E0>0 where E0 is constant. If at t=0, de-Broglie wavelength is λ0=hmv0, then its de-Broglie wavelength after time t is given by

Options:
A λ0
B λ01+eE0tmv0
C λ0t
D λ01+eE0tmv0

The correct answer is:
λ01+eE0tmv0

Solution:
Initial de-Broglie wavelength of electron,

λ0=hmv0 ....(i)

Force on electron in electric field,

F=-eE=-e-E0i^=eE0i^

Acceleration of electron, a=Fm=eE0i^m
Velocity of electron after time t,

v=v0i^+eE0i^mt=v0+eE0mti^=v01+eE0mv0ti^

de-Broglie wavelength associated with electron at time t is

λ=hmv

⇒λ=hmv01+eE0mv0t=λ01+eE0mv0t

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (27 Jul Shift 1)
An electron (mass m ) with initial velocity v→=v0i^+v0j^ is in an electric filed E→=-E0k^. If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by:

Options:
A λ021+e2E2t2m2v02
B λ01+e2E2t2m2v02
C λ01+e2E2t22m2v02
D λ02+e2E2t2m2v02

The correct answer is:
λ01+e2E2t22m2v02

Solution:
Initial velocity of electron,

v=v0i^+v0j^⇒u=v02+v02=v02

Initial de Broglie wavelength associated with electron,

λ0=hp=hmv02

now,

v=u+at
Velocity as a function of time =v0i^+v0j^+eE0mtk^
so wavelength λ=hm2v02+e2E02m2t2
λ=λ01+e2E022m2v02t2

Topic: Physics, Dual Nature of Matter, JEE Main
An electron (mass m ) with initial velocity v→=v0i^+v0j^ is in an electric filed E→=-E0k^. If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by:

Options:
A λ021+e2E2t2m2v02
B λ01+e2E2t2m2v02
C λ01+e2E2t22m2v02
D λ02+e2E2t2m2v02

The correct answer is:
λ01+e2E2t22m2v02

Solution:
Initial velocity of electron,

v=v0i^+v0j^⇒u=v02+v02=v02

Initial de Broglie wavelength associated with electron,

λ0=hp=hmv02

now,

v=u+at
Velocity as a function of time =v0i^+v0j^+eE0mtk^
so wavelength λ=hm2v02+e2E02m2t2
λ=λ01+e2E022m2v02t2

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (08 Jan Shift 2)
An electron microscope is superior to an optical microscope in

Options:
A Having better resolving power
B Being easy to handle
C Low cost
D Quickness of observation

The correct answer is:
Having better resolving power

Solution:
)In electron microscope, electron beam $(\lambda \approx 1 A)$ is used so it's R.P. is approx. 5000 times more than that of ordinary microscope $(\lambda \approx 5000 \mathrm{~A})$.

Topic: Physics, Ray Optics, JEE Main
An electron moves through a uniform magnetic field B→=B0i^+2B0j^ T. At a particular instant of time, the velocity of electron is u→=3i^+5j^ m s-1. If the magnetic force acting on electron is F→=5ek^ N, where e is the charge of electron, then the value of B0 is ____ T.

Options:

The correct answer is:
5

Solution:
Force acting on a charged particle under the magnetic field is given by, F→=qv→×B→.

⇒5ek⏜=e3i⏜+5j⏜×B0i⏜+2B0j⏜

⇒5ek⏜=e6B0k⏜-5B0k⏜

⇒B0=5 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (31 Jan Shift 1)
An electron moves with speed $2 \times 10^5 \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in the presence of a magnetic field of induction $\mathbf{B}=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ (in tesla). The magnitude of the force experienced by the electron in newtons is
$\left(\right.$ Charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A $1.18 \times 10^{-13}$
B $1.28 \times 10^{-13}$
C $1.6 \times 10^{-13}$
D $1.72 \times 10^{-13}$

The correct answer is:
$1.6 \times 10^{-13}$

Solution:
$\begin{aligned} & \mathbf{v}=2 \times 10^5 \mathrm{~m} / \mathrm{s} \text { (along } X \text {-axis) } \\ & =\left(2 \times 10^4 \hat{\mathbf{i}}\right) \mathrm{m} / \mathrm{s} \\ & \mathbf{B}=(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \mathrm{T} \\ & \text { Force } \quad \mathbf{F}=q(\mathbf{v} \times \mathbf{B}) \\ & \text { For electron } q=1.6 \times 10^{-19} \mathrm{C} \\ & \mathbf{V} \times \mathbf{B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 \times 10^5 & 0 & 0 \\ 1 & 4 & -3\end{array}\right| \\ & =\hat{\mathbf{i}}[(0 \times-3)-(4 \times 10)] \\ & -\hat{\mathbf{j}}\left[\left(2 \times 10^5 \times-3\right)-(1 \times 0)\right] \\ & +\hat{\mathbf{k}}\left[\left(2 \times 10^5 \times 4\right)-(1 \times 0)\right] \\ & =6 \times 10^5 \hat{\mathbf{j}}+8 \times 10^5 \hat{\mathbf{k}} \\ & \mathbf{F}=1.6 \times 10^{-19} \sqrt{\left(6 \times 10^5\right)^2+\left(8 \times 10^5\right)^2}, \\ & =1.6 \times 10^{-19} \times 10 \times 10^5 \\ & =1.6 \times 10^{-18} \mathrm{~N} \\ & \end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron, moving along the x- axis with an initial energy of 100 eV, enters a region of magnetic field B→=1.5×10-3 Tk^ at S (see figure). The field extends between x=0 and x=2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is:
(electron’s charge 1.6×10-19 C , mass of electron =9.1×10-31 kg )

Options:
A 1.22 cm
B 12.87 cm
C 11.65 cm
D 2.25 cm

The correct answer is:
12.87 cm

Solution:
R=mVqB
=2m(K.E)qB
mV=2m K.E
=2×9.1×10-31×100×1.6×10-191.6×10-19×1.5×10-3
=2.248 cm≈ 2.25 cm
sin⁡θ=22.248=0.89
PA=R1-cos⁡θ=1.22 cm
AB=6tan⁡θ=11.71 cm
d=PA+AB=12.93 cm

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (12 Apr Shift 2)