datavorous/entrance-exam-dataset huggingface dataset:
An electromagnetic wave of frequency 1× 10 14  Hz is propagating along z -axis. The amplitude of the electric field is 4 V/m. If ε0=8.8×10-12 C2/N-m2, then the average energy density of electric field will be

Options:
A 35.2Ã 10 â12 â J/m 3
B 35.2× 10 −10   J/m 3
C 35.2× 10 −11   J/m 3
D 35.2× 10 −13   J/m 3

The correct answer is:
35.2× 10 −12   J/m 3

Solution:
fâ¡ = 1 0 1 4 Hz
E 0 =4 V/m
â 0 = 8.8 Ã 1 0 - 1 2 C 2 N - m 2
Energy density of electric field
= 1 2 (Total energy density)
= 1 2 Â· 1 2 â 0 E 2
= 1 2 . 1 2 8.8 Ã 1 0 - 1 2 Ã 4 2
= 1 2 · 1 2 ×16×8.8× 10 −12   J/m 3
=35.2× 10 −12   J/m 3

Topic: Physics, Electromagnetic Waves, JEE Main
An electromagnetic wave of frequency $2 \mathrm{MHz}$ propagates from vacuum to a non-magnetic medium of relative permittivity 9 . Then its' wavelength

Options:
A increases by 100 m
B increases by 50 m
C decreases by 50 mm
D decreases by 100 m

The correct answer is:
decreases by 100 m

Solution:
In vacuum wavelength,
$\lambda_0=\frac{c}{f_0}=\frac{3 \times 10^8}{2 \times 10^6}=1.5 \times 10^2=150 \mathrm{~m}$
In medium, speed of wave is
$v=\frac{\mathcal{c}}{\sqrt{\varepsilon_r \mu_r}} \approx \frac{\mathcal{c}}{\sqrt{\varepsilon_r}}$
(as medium is non-magnetic)
$\therefore \quad v=\frac{\mathcal{c}}{\sqrt{9}}=\frac{\mathcal{c}}{3}=1 \times 10^8 \mathrm{~ms}^{-1}$
Frequency remaining same, wavelength in medium is $\lambda=\frac{v}{f_0}=\frac{10^8}{2 \times 10^6}=50 \mathrm{~m}$
$\therefore$ Wavelength decreases by $100 \mathrm{~m}$.

Topic: Physics, Electromagnetic Waves, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _________ ×10-2 cm.

Options:

The correct answer is:
667

Solution:
λ in vacuum =cf=3×1083×109=0.1 m

∴ λ in medium =0.1μ

Where refractive index

μ=μrεr

Assuming non-magnetic material μr=1

∴ μ=2.25=1.5

λm=0.11.5=115 m=6.67 cm

=667×10-2 cm

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (24 Feb Shift 2)
An electromagnetic wave of frequency 5GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is _______×107 m s-1.

Options:

The correct answer is:
15

Solution:
Given : Frequency of wave f=5GHz

=5×109 Hz

Relative permittivity, ϵr=2 and Relative permeability, μr=2 Since speed of light in a medium is given by,

v=1μ∈=1μrμ0·ϵr∈0

v=1μrϵf1μ0ϵ0=Cμrϵr

Where C is speed of light is vacuum

∴ v=3×1084=30×1072 m s-1

=15×107 m s-1

∴ Ans. is 15

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (24 Feb Shift 1)
An electromagnetic wave of frequency Ï=3.0 MHz passes from vacuum into a dielectric medium with permittivity Îµ=4.0. Then

Options:
A wavelength is doubled and the frequency remains unchanged
B wavelength is doubled and frequency becomes half
C wavelength is halved and frequency remains unchanged
D wavelength and frequency both remain unchanged

The correct answer is:
wavelength is halved and frequency remains unchanged

Solution:
During propagation of a wave from one medium to another, the frequency remains constant and wavelength changes.
μ=εε0=4=2
Since μ∝1λ
∴ Wavelength is halved
Hence option (wavelength is halved and frequency remains unchanged) holds good.

Topic: Physics, Electromagnetic Waves, NEET
An electromagnetic wave of frequency 3.0 MHz passes from vacuum into a non-magnetic medium with permittivity, $\epsilon=16 \epsilon_0$. Where, $\epsilon_0$ is the free space permittivity. The change in wavelength is

Options:
A $-75 \mathrm{~m}$
B $+75 \mathrm{~m}$
C $-50 \mathrm{~m}$
D $+50 \mathrm{~m}$

The correct answer is:
$-75 \mathrm{~m}$

Solution:
Given, frequency of EM wave,
$\begin{aligned}
f & =3 \mathrm{MHz} \\
& =3 \times 10^6 \mathrm{~Hz}
\end{aligned}$
and permittivity of non-magnetic medium,
$\varepsilon=16 \varepsilon_0$
Wavelength of EM wave,
$\begin{aligned}
& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{3 \times 10^6} \\
& \lambda=100 \mathrm{~m}
\end{aligned}$
$\lambda=100 \mathrm{~m}$
Velocity of electromagnetic (EM) into non-magnetic material,
$\begin{aligned}
v & =\frac{c}{\sqrt{\varepsilon_{\varphi}}}=\frac{3 \times 10^8}{\sqrt{16}} \quad\left[\because \varepsilon_{\varphi}=\frac{\varepsilon}{\varepsilon_0}=16\right] \\
& =\frac{3}{4} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\therefore$ Wavelength, $\lambda^{\prime}=\frac{v}{f}=\frac{\frac{3}{4} \times 10^8}{3 \times 10^6}=25 \mathrm{~m}$
$\therefore$ Change in wavelength $=\lambda^{\prime}-\lambda$
$=25-100=-75 \mathrm{~m}$

Topic: Physics, Electromagnetic Waves, JEE Main
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with permittivity $\varepsilon=4.0$. Then

Options:
A
wavelength is doubled and the frequency remains unchanged

B
wavelength is doubled and frequency becomes half

C
wavelength is halved and frequency remains unchanged

D
wavelength and frequency both remain unchanged.

The correct answer is:
wavelength is halved and frequency remains unchanged

Solution:
Refractive index, $\mu=\sqrt{\frac{\varepsilon}{\varepsilon_0}}=2$
Speed and wavelength of wave will becomes half, the frequency remaining unchanged (frequency of a wave depends on the source as due to refraction, it is assumed that the energy is conserved. hv remains the same)

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2004
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\varepsilon=4.0$ Then:

Options:
A Wavelength is double and frequency become half.
B Wavelength is halved and frequency remains unchanged.
C Wavelength and frequency both remain unchanged.
D Wavelength is doubled and frequency unchanged.

The correct answer is:
Wavelength is halved and frequency remains unchanged.

Solution:
Velocity of electromagnetic wave in vacuum.
$$
C=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=V_{\mathrm{vacum}}
$$
Velocity of electromagnetic wave in medium
$$
V_{m e d}=\frac{1}{\sqrt{\mu_0 \mu_0 \varepsilon_0 c_{n r}}}=\frac{\mathrm{C}}{\sqrt{\mu_r \varepsilon_r}}
$$
$\mu_0$ and $\varepsilon_0$ is in vacuum and $\mu \mathrm{r} \varepsilon_0$ is in medium.
For dielectric medium $r=1$
$$
\begin{aligned}
& \therefore V_{\text {med }}=\frac{C}{\sqrt{\varepsilon_r}} \text { Given } \varepsilon_0=4 \\
& \therefore V_{\text {med }}=\frac{C}{\sqrt{4}}=\frac{\mathrm{C}}{2}
\end{aligned}
$$
$\therefore$ Wavelength of the wave in medium $\lambda_{\text {med }}=\frac{V_{\text {med }}}{V}=\frac{C}{2 V}=$ $\frac{\lambda_{\text {vacum }}}{2}$

Topic: Physics, Electromagnetic Waves, JEE Main
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\varepsilon=4.0$ Then:

Options:
A Wavelength is double and frequency become half.
B Wavelength is halved and frequency remains unchanged.
C Wavelength and frequency both remain unchanged.
D Wavelength is doubled and frequency unchanged.

The correct answer is:
Wavelength is halved and frequency remains unchanged.

Solution:
Velocity of electromagnetic wave in vacuum.
$$
C=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=V_{\mathrm{vacum}}
$$
Velocity of electromagnetic wave in medium
$$
V_{m e d}=\frac{1}{\sqrt{\mu_0 \mu_0 \varepsilon_0 c_{n r}}}=\frac{\mathrm{C}}{\sqrt{\mu_r \varepsilon_r}}
$$
$\mu_0$ and $\varepsilon_0$ is in vacuum and $\mu \mathrm{r} \varepsilon_0$ is in medium.
For dielectric medium $r=1$
$$
\begin{aligned}
& \therefore V_{\text {med }}=\frac{C}{\sqrt{\varepsilon_r}} \text { Given } \varepsilon_0=4 \\
& \therefore V_{\text {med }}=\frac{C}{\sqrt{4}}=\frac{\mathrm{C}}{2}
\end{aligned}
$$
$\therefore$ Wavelength of the wave in medium $\lambda_{\text {med }}=\frac{V_{\text {med }}}{V}=\frac{C}{2 V}=$ $\frac{\lambda_{\text {vacum }}}{2}$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2013 (Karnataka)
An electromagnetic wave passes through space and its equation is given by $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$
where $\mathrm{E}$ is electric field. Energy density of electromagnetic wave in space is

Options:
A $\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
B $\frac{1}{4} \varepsilon_{0} \mathrm{E}_{0}^{2}$
C $\varepsilon_{0} \mathrm{E}_{0}^{2}$
D $2 \varepsilon_{0} \mathrm{E}_{0}^{2}$

The correct answer is:
$\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$

Solution:
Energy density
$$
=\varepsilon_{0} \mathrm{E}_{\mathrm{rm}}^{2}=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}
$$

Topic: Physics, Electromagnetic Waves, JEE Main
An electromagnetic wave travelling in the x- direction has frequency of 2×1014 H z and electric field amplitude of 27 V m–1 oscillates in Y-direction. From the options given below, which one describes the magnetic field for this wave?

Options:
A B→x, t=(9×10-8 T)j^ sin1.5×10-6 x-2×1014t
B B→x, t=(9×10-8T)i^sin2π1.5×10-8 x-2×1014t
C B → x,t=3×10-8Tj^sin2πx1.5×10-8-2×1014t
D B → x,t=9×10-8Tk^sin2πx1.5×10-6-2×1014t

The correct answer is:
B → x,t=9×10-8Tk^sin2πx1.5×10-6-2×1014t

Solution:
When undefined
Then B = B0sin⁡ ( kx−ωt )
Of light in travelling along i^ then B→ in either along j ^ or k ^ .
∴ speed of light C= E0B0 ⇒B0= E0C
⇒ B0= 273×108=9 ×10-8 T
also, ω=2π f=2π ×2 ×1014=4 π ×1014
k= ω c = 4π× 10 14 3× 10 8
=( 2π 1.5× 10 −6 )
Looking into the option the correct
Answer is B→=9×10-8sin⁡2πx1.5×10-6-2×1014tk^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2015 (10 Apr Online)
An electromagnetic wave with frequency $\omega$ and wavelength $\lambda$ travels in the $+y$ direction. Its magnetic field is along $+x$-axis. The vector equation for the associated electric field (of amplitude $E_0$ ) is

Options:
A
$\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{x}$

B
$\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{x}$

C
$\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}$

D
$\vec{E}=-E_0 \cos \left(\omega t+\frac{2 \pi}{\lambda} y\right) \hat{z}$

The correct answer is:
$\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}$

Solution:
In an electromagnetic wave electric field and magnetic field are perpendicular to the direction of propagation of wave. The vector equation for the electric field is
$$
\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}
$$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2012 (19 May Online)
An electron, a doubly ionized helium ion ($He^{++}$) and proton are having the same kinetic energy. The relation between their respective de-Broglie wavelength λe*λHe++ and λp is :

Options:
A $\lambda_e>\lambda_{{He}^{++}}>\lambda_p$
B $\lambda_e<\lambda_{{He}^{++}}==\lambda_p$
C $\lambda_e>\lambda_p>\lambda_{{He}^{++}}$
D $\lambda_e<\lambda_p<\lambda_{{He}^{++}}$

The correct answer is:
$\lambda_e>\lambda_p>\lambda_{{He}^{++}}$

Solution:
$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m k}}$
$\lambda_{\mathrm{e}}: \lambda_{\mathrm{p}}: \lambda_{{He}^{++}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}: \frac{1}{\sqrt{\mathrm{He}^{++}}}$
$\frac{1}{\sqrt{9.1 \times 10^{-31}}}: \frac{1}{\sqrt{1.6 \times 10^{-27}}}: \frac{1}{\sqrt{4 \times 1.6 \times 10^{-27}}}=84: 2: 1$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (06 Sep Shift 1)
An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, rα respectively in a uniform magnetic field B. The relation between re, rp, rα is:

Options:
A re<rα<rp
B re>rP=rα
C re<rp=rα
D re<rp<rα

The correct answer is:
re<rp=rα

Solution:
r=mVqB=2mKqB⇒r∝mq. For electron, re=kmee Where, me is the mass of an electron. m=the mass of a proton. For alpha particles, rα=k4m2e=kme, q=charge on proton. For proton, rp=kme, so, re<rp=rα.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (08 Apr)
An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50%. The value of V1V2 is equal to :

Options:
A 3
B 94
C 32
D 4

The correct answer is:
94

Solution:
Let initial wavelength be λ, then after 50% increase wavelength will become 1.5λ.

Now, KE=P22m=eV & P=hλ

Therefore, eV=hλ22m.

So we can write

eV1=hλ22m ...Eq(1) and

eV2=h1.5λ22m ...Eq(2)

From both equations, we get

V1V2=1.52=94

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (30 Jan Shift 2)
An electron beam can undergo diffraction by crystals. The potential of 'V' volt should a beam of electrons be accelerated so that its wavelength becomes equal to 1.0 Ã . The value of 'V' is
Given: h 2 m e Ãe =3Ã 10 â18 J 2 s 2 kg â1 coulomb â1

Options:

The correct answer is:
150

Solution:
We know that
12mu2=eV
and Î»=hmu or u=hmÎ» or u2=h2m2Î»2
12mÃh2m2Î»2=eV
or V =12mÃh2m2Î»2Ãe=12Ãh2mÎ»2Ãe
Substituting the values, we get
V=12Ã6.62Ã10-3429.108Ã10-31Ã1Ã10-102Ã1.602Ã10-19=150 volt

Topic: Chemistry, Structure of Atom, JEE Main
An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of logλmin with logV is correctly represented in :

Options:
A

B

C

D

The correct answer is:

Solution:
K.E.=eV,

KE=The kinetic energy of electron
The maximum energy of photon=K.E.
eV=hCλmin

⇒λmin V=hCe=constant
Taken log in both sides
lnλmin+lnV=lnconstant
⇒lnλmin=-lnV+ln(constant)
y=-ax+c the straight line of negative slope.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2017 (02 Apr)
An electron beam travels with a velocity of $1.6 \times 10^7 \mathrm{~ms}^{-1}$ perpendicularly to magnetic field of intensity $0.1 \mathrm{~T}$. The radius of the path of the electron beam $\left(m_e=9 \times 10^{-31} \mathrm{~kg}\right.$ )

Options:
A $9 \times 10^{-5} \mathrm{~m}$
B $9 \times 10^{-2} \mathrm{~m}$
C $9 \times 10^{-4} \mathrm{~m}$
D $9 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$9 \times 10^{-4} \mathrm{~m}$

Solution:
In a perpendicular magnetic field, the radius of circular path travelled by electron beam is
$\begin{aligned} r & =\frac{m v}{e B} \\ \therefore \quad r & =\frac{9 \times 10^{-31} \times 1.6 \times 10^7}{1.6 \times 10^{-19} \times 0.1} \\ & =9 \times 10^{-4} \mathrm{~m}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron collides with a free molecules initially in its ground state. The collision leaves the molecules in an excited state that is metastable and does not decay to the ground state by radiation. Let $\mathrm{K}$ be the sum of the initial kinetic energies of the electron and the molecule, and $\vec{P}$ the sum of their initial momenta. Let $\mathrm{K}^{\prime}$ and $\vec{P}$ 'represent the same physical quantities after the collision. Then -

Options:
A $K=K^{\prime}, \vec{P}=\vec{P}^{\prime}$
B $K^{\prime} < K, \vec{P}=\vec{P}^{\prime}$
C $K=K^{\prime}, \vec{P} \neq \vec{P}^{\prime}$
D $K^{\prime} < K, \vec{P} \neq \vec{P}^{\prime}$

The correct answer is:
$K^{\prime} < K, \vec{P}=\vec{P}^{\prime}$

Solution:
Collision of e $^{-}$lead to excitation of molecules so Collision is inelastic
$\therefore \mathrm{K}^{\prime} < \mathrm{K}$ and loss of kinetic energy go for excitation of molecules. Momentum remain conserved during collision.
$\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{P}^{\prime}}$

Topic: Physics, Center of Mass Momentum and Collision, KVPY, KVPY 2011 (SB/SX)
An electron enters a chamber in which a uniform magnetic field is present as shown. Ignore gravity.
During its motion inside the chamber

Options:
A the force on the electron remains constant
B the kinetic energy of the electron remains constant
C the momentum of the electron remains constant
D the speed of the electron increases at a uniform rate

The correct answer is:
the kinetic energy of the electron remains constant

Solution:
$\because$ work done $=0$
Hence kinetic energy $=$ constant

Topic: Physics, Magnetic Effects of Current, KVPY, KVPY 2013 (SB/SX)
An electron enters a magnetic field along perpendicular direction. Following quantity will remain constant

Options:
A momentum
B kinetic energy
C velocity
D all of the above

The correct answer is:
kinetic energy

Solution:
Work done by force of magnetic field on moving charge is zero therefore kinetic energy and speed remains constant while direction of velocity and hence the momentum may change.

Topic: Physics, Magnetic Effects of Current, NEET
An electron enters an electric field having
intensity $\mathbf{E}=3 \overrightarrow{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \cdot \mathbf{V} \mathbf{m}^{-1}$ and magnetic
field having induction $\mathrm{B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ T with a velocity $\mathbf{v}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ ms $^{-1}$. The magnitude of the force acting on the electron is (Given, $\left.e=-1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A $2.02 \times 10^{-18} \mathrm{N}$
B $5.16 \times 10^{-16} \mathrm{N}$
C $3.72 \times 10^{-17} \mathrm{N}$
D None of the above

The correct answer is:
None of the above

Solution:
Given, $E=3 i+6 j+2 \hat{k} \vee / m$
$$
\begin{array}{l}
B=2 \hat{i}+3 \hat{j} T \\
v=2 \hat{i}+3 \hat{j} m / s \\
e=-1.6 \times 10^{-10} C
\end{array}
$$
The magnitude of electric field
$$
\begin{aligned}
|E| &=\sqrt{(3)^{2}+(6)^{2}+(2)^{2}} \\
&=\sqrt{9+36+4} \\
&=\sqrt{49}=7
\end{aligned}
$$
The force acting on electron due to electric field
$$
F=q E
$$
$$
\begin{aligned}
|F| &=|q||\mathbf{E}| \\
&=1.6 \times 10^{-19} \times 7 \mathrm{N} \\
&=11.2 \times 10^{-19} \mathrm{N} \\
&=1.12 \times 10^{-18} \mathrm{N}
\end{aligned}
$$
Alternate Method
B $\|$ v Magnetic field is parallel to velocily of electron. $\mathrm{v} \| \mathrm{B}$ Magnetic force
$$
\begin{aligned}
F_{m} &=q(\mathbf{v} \times \mathbf{B}) \\
&=q v B \sin \theta \\
&=q v B \sin 0 \\
&=0
\end{aligned}
$$
$\therefore$ Net force $\quad F=F_{e}+F_{m}$
$$
\begin{array}{l}
=\left(1.12 \times 10^{-18}+0\right) \mathrm{N} \\
=1.12 \times 10^{-18} \mathrm{N}
\end{array}
$$
Hence, none of the given options is correct.

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn, λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let ∧n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

Options:
A ∧n2≈λ
B ∧n≈A+Bλn2
C ∧n≈A+Bλn
D ∧n2≈A+Bλn2

The correct answer is:
∧n≈A+Bλn2

Solution:
2πr=n⋋n
⋋n=2πrn
∴r=ron2
⋋n=2πn ron2=n2πro …..i
1∧​n=R112-1n2
∧​n=1R1+1n2-1
∧​n=1R1+1n2 ………..ii
from (i) and (ii)
∧​n=1R1+4n2ro2⋋n2=A+B⋋n2

Topic: Physics, Atomic Physics, JEE Main
An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λn, λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let ∧n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

Options:
A ∧n2≈λ
B ∧n≈A+Bλn2
C ∧n≈A+Bλn
D ∧n2≈A+Bλn2

The correct answer is:
∧n≈A+Bλn2

Solution:
2πr=n⋋n
⋋n=2πrn
∴r=ron2
⋋n=2πn ron2=n2πro …..i
1∧​n=R112-1n2
∧​n=1R1+1n2-1
∧​n=1R1+1n2 ………..ii
from (i) and (ii)
∧​n=1R1+4n2ro2⋋n2=A+B⋋n2

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (08 Apr)
An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns/length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed v. If the electron does not hit the surface of the solenoid, maximum possible value of v is (all symbols have their standard meaning):

Options:
A eμ0nIRm
B eμ0nIR2m
C eμ0nIR4m
D 2eμ0nIRm

The correct answer is:
eμ0nIRm

Solution:
Magnetic Field Inside Solenoid is B=μ0nI

Velocity ν has given to the electron

So, Magnetic Force

F=qνB

as ν and B are perpendicular.

In a magnetic field, charge will move on a circular path. Here, If radius is just less than or equal to radius of solenoid then the electron does not hit the surface.

F=qνB=mν2R⇒mν=qBRas q=eν=eBRmSo,V=eμ0nIRm

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (09 Jan Shift 2)
An electron in a circular orbit of radius $0.05 \mathrm{nm}$ performs $10^{16}$ revolutions per second. The maghetic moment due to this rotation of electron is (in $\mathrm{Am}^{2}$ )

Options:
A $2.16 \times 10^{-23}$
B $3.21 \times 10^{-22}$
C $3.21 \times 10^{-24}$
D $1.26 \times 10^{-23}$

The correct answer is:
$1.26 \times 10^{-23}$

Solution:
Given, $r=0.05 \mathrm{nm}=0.05 \times 10^{-9} \mathrm{m}$
$$
\begin{array}{l}
\qquad\left(\because 1 \mathrm{nm}=10^{-9} \mathrm{m}\right) \\
n=10^{16} \text { revolutions } \\
e=1.6 \times 10^{-19} \mathrm{C}
\end{array}
$$
We know that
The magnetic moment $M=A L$
$$
\begin{aligned}
M &=\pi r^{2} \times n e \\
M &=3.14 \times\left(0.05 \times 10^{-9}\right)^{2} \times 10^{16} \times 1.6 \times 10^{-19} \\
M &=0.1256 \times 10^{-18} \times 10^{16} \times 10^{-19} \\
\text {or } M &=1.26 \times 10^{-23} \mathrm{Am}^{2}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
An electron in a hydrogen atom is in an orbit from which there can be a maximum of six transitions. Another electron in a different hydrogen atom is in an orbit from which there can be a maximum of three transitions. The ratio of the velocities of the electrons in these two orbits is

Options:
A 12
B 23
C 54
D 34

The correct answer is:
34

Solution:
Number of spectral lines obtained due to transition of electrons from nth orbit to lower orbit is,
N= n(n-1)2
For case - 1, N = 6 = n1(n1-1)2
⇒ n1=4
For case - 2, N =3 = n2(n2-1)2
⇒ n2=3
The velocity of electron in a hydrogen atom in nth orbit
vn∝1n
vnv′n=n2n1
⇒ n6n3= 34

Topic: Physics, Atomic Physics, JEE Main
An electron in a hydrogen atom jumps from second Bohr orbit to ground state and the energy difference of the two states is radiated in the form of photons. These are then allowed to fall on a metal surface having a work-function equal to 4.2 eV, then the stopping potential is [Energy of electron in nth orbit =-13.6n2 eV]

Options:
A 2 V
B 4 V
C 6 V
D 8 V

The correct answer is:
6 V

Solution:
E=13.61-13.622
E=13.64-14=13.6×34
∴ E=10.2 eV =hν
i.e. hν=10.2eV
hν=ϕ0+eVs
∴ 10.2 eV=4.2 eV+eVs
∴ 6 eV=eVs
∴ Vs=6 V

Topic: Physics, Dual Nature of Matter, JEE Main
An electron in a hydrogen atom makes a transition from n=2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom which was already in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the lithium-ion for the process is

Options:
A 2
B 4
C 5
D 3

The correct answer is:
4

Solution:
Let us assume that the electron in the lithium-ion is knocked off from the nth energy level, then

RZH2112−122=RZLi2n2

R12×34=R321n2

34=9n2
n=12≈3.5

Hence the least quantum number must be 4.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2016 (09 Apr Online)
An electron in a hydrogen atom makes a transition from n=2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom which was already in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the lithium-ion for the process is

Options:
A 2
B 4
C 5
D 3

The correct answer is:
4

Solution:
Let us assume that the electron in the lithium-ion is knocked off from the nth energy level, then

RZH2112−122=RZLi2n2

R12×34=R321n2

34=9n2
n=12≈3.5

Hence the least quantum number must be 4.

Topic: Physics, Atomic Physics, JEE Main