datavorous/entrance-exam-dataset huggingface dataset:
An electric kettle has two heating coils. When one of the coils is connected to a.c. source, the water in the kettle boils in 10 minute. When the other coil is used the water boils in 40 minute. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be:

Options:
A 8 minute
B 4 minute
C 25 minute
D 15 minute

The correct answer is:
8 minute

Solution:
Here $Q=\frac{V^2}{R_1} \times t_1=\frac{V^2}{R_2} \times t_2$
$$
\begin{aligned}
& =\frac{V^2}{R} \times t \\
& \therefore \quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \quad \frac{Q}{V^2 t}=\frac{Q}{V^2 t_1}+\frac{Q}{V^2 t_2} \\
& \Rightarrow \quad \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_4} \\
& \Rightarrow \quad t=\frac{t_1 t_2}{t_1+t_2} \\
& =\frac{10 \times 40}{10+40}=8 \mathrm{~min} \\
&
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2003
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$

Options:
A $6.3 \mathrm{~min}$
B $8.4 \mathrm{~min}$
C $12.6 \mathrm{~min}$
D $4.2 \mathrm{~min}$

The correct answer is:
$6.3 \mathrm{~min}$

Solution:
Key Idea : Heat eyolved due to Joule's effect is used up in boiling water.
As per key idea
$\begin{aligned}
V I t & =m s \Delta t \\
\text{or } t & =\frac{m s \Delta t}{V}
\end{aligned}$
Putting under given values
$\begin{gathered}
I=4 \mathrm{~A}, V=220 \text { volt }, \quad m=1 \mathrm{~kg}, \\
\Delta t=(100-20)^{\circ} \mathrm{C}, s=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \\
\therefore \quad t=\frac{1 \times 4200 \times 80}{220 \times 4}=6.3 \mathrm{~min}
\end{gathered}$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2008 (Screening)
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 m s-1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : g=10 m s-2

Options:
A 20000
B 34500
C 23500
D 23000

The correct answer is:
34500

Solution:
Since the speed is constant the motor has to give equal and opposite upward force.

∴ F=3000N+2000×10 =23000 N

P=FV=23000×1.5 =34500 W

Topic: Physics, Work Power Energy, NEET, NEET 2022 (Phase 1)
An electric motor which is loaded has an effective resistance of 30 Î© and an inductive reactance of 40 Î©. If the motor is powered by a source with a maximum voltage of 420 V, the maximum current is

Options:
A 6 A
B 8.4 A
C 10 A
D 12 A

The correct answer is:
8.4 A

Solution:
Given,
Effective resistance, R=30 Î©
Inductive reactance, XL=40 Î©
The effective impedance
Z=R2+XL2
=302+402
=900+1600
=2500
=50Î©
The maximum current,
I=VZ
=42050
=8.4 A

Topic: Physics, Alternating Current, NEET
An electrical cable of copper has just one wire of radius 9 mm and its resistance is 5 Î©. This single copper wire of cable is replaced by 6 different well-insulated copper wires, each of radius 3 mm. The total resistance of cable will now be equal to

Options:
A 7.5 Î©
B 45Â Î©
C 90 Î©
D 270 Î©

The correct answer is:
7.5 Î©

Solution:
R â1r2
âµRadius 9 mmâ¶3 mm
âResistance 5 Î©â¶ 45 Î©
Result of six parallel wires=456=7.5 Î©

Topic: Physics, Current Electricity, NEET
An electrochemical cell is set up as: Pt; H2 (1 atm)|HCl(0.1 M)|| CH3COOH (0.1 M)| H2 (1 atm); Pt. The e.m.f. of this cell will not be zero, because

Options:
A acids used in two compartments are different
B e.m.f. depends on molarities of acids used
C the temperature is constant
D pH of 0.1 M HCl and 0.1 M CH3COOH is not same

The correct answer is:
pH of 0.1 M HCl and 0.1 M CH3COOH is not same

Solution:
Since it is a concentration cell and the concentration of ions in two electrolyte solutions (HCl and CH3COOH ) are different, therefore e.m.f. of this cell will not be zero.

Topic: Chemistry, Electrochemistry, NEET
An electrochemical cell stops working after sometime because

Options:
A one of the electrodes is eaten away
B electrode potentials of both the electrodes go on increasing
C electrode potential of both the electrodes go on decreasing
D electrode potentials of the two electrodes become equal in magnitude

The correct answer is:
electrode potentials of the two electrodes become equal in magnitude

Solution:
An electrochemical cell stops working after sometime because at equilibrium, electrodes potential of the two electrodes become equal in magnitude.

Topic: Chemistry, Electrochemistry, NEET
An electrolyte cell contains a solution of Ag2SO4 and have platium electrodes. A current is passed until 1.6 g of O2 has been liberated at anode. The amount of silver deposited at cathode would be

Options:
A 107.88 g
B 0.8 g
C 1.6 g
D 21.60 g

The correct answer is:
21.60 g

Solution:
O2--2e→12O2
Equivalent weight of oxygen =162=8
8gO2=1 g equivalent
1.6 g O2=168=0.2 g equivalent O2
1 g equivalent of Ag = 108 g Ag
0.2 g equivalent of Ag = 0.2×108 g Ag
= 21.60 g Ag

Topic: Chemistry, Electrochemistry, NEET
An electrolytic cell contains a solution of Ag2SO4 and has platinum electrodes. A current is passed until 1.6 g of O2 has been liberated at anode. The amount of silver deposited at cathode will be

Options:
A 108 g
B 1.6 g
C 0.8 g
D 21.60 g

The correct answer is:
21.60 g

Solution:
W1W2=E1E2
1.6W2=8108
W2=21.6g

Topic: Chemistry, Electrochemistry, NEET
An electrolytic cell is constructed for preparing hydrogen. For an average current of 2 ampere in the circuit, the time required to produce 450 mL of hydrogen at NTP is approximately

Options:
A 1/2 hour
B 1 hour
C 2 hours
D 5 hours

The correct answer is:
1/2 hour

Solution:
22400 mL H 2 at STP =â1âmolâ H 2
So 450âmlâ H 2 at STP
=â 450âmL 22400âmL Ã1âmolâ H 2
=0.02 mol
Electrode reaction
2H++2e-2FâH21mol
The liberation of 1âmolâ H 2 requires =2Ã96500âC
So the liberation of 0.02âmolâ H 2 requires =2Ã96500Ã0.02âC
=3860 C
Time required to produce 0.02âmolâ H 2 ,âtâ=â Q i
=â 3860 2A â=â1930âsâââ0.503âhour
â0.5âhour

Topic: Chemistry, Electrochemistry, NEET
An electromagnetic wave is propagating along the +z-axis. Direction of electric and the magnetic field of this wave can be represented by

Options:
A E→=E0i^,  B→=B0j^
B E→=E0k^,  B→=B0i^
C E→=E0j^,  B→=B0i^
D Eâ=E0j^,ââBâ=B0k^

The correct answer is:
E→=E0i^,  B→=B0j^

Solution:
The electromagnetic wave is propagating along the +z axis.
Since the electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation of wave.
Also, Eâ¡ â Ã Bâ¡ â gives the direction of wave propagation.
Eâ¡ â = Eâ¡ 0 iâ¡ ^ , B = Bâ¡ 0 jâ¡ ^ âµ iâ¡ ^ Ã jâ¡ ^ = k ^

Topic: Physics, Electromagnetic Waves, NEET
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, NEET
An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If m mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then :

Options:
A λ=2 mhcλd2
B λd=2mchλ2
C λ=2mchλd2
D λ=2hmcλd2

The correct answer is:
λ=2mchλd2

Solution:
From Einstein's equation,

Eincident=ϕ+kEmax

hf=0+12mv2

hcλ=p22m⇒p=2mhcλ

λd=hp=h2mhcλ=λh2mc

⇒λ=2mchλd2

Topic: Physics, Dual Nature of Matter, NEET, NEET 2021
An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelength are Î»e, Î»n and Î»Î± respectively. Which statement is correct about their de-Broglie wavelengths?

Options:
A λe>λn>λα
B λe<λn>λα
C λe<λn<λα
D λe>λn<λα

The correct answer is:
λe>λn>λα

Solution:
According to de - Broglie theory, the wavelength
λ=hp=h2mk
λ∝1m (for same kinetic energy)
The wavelength is inversely proportional to the root of mass of the particle.
As me<mn<mα
So, λe>λn>λα

Topic: Physics, Dual Nature of Matter, NEET
An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively Re, Rp, Rd and RÎ±. If follows that

Options:
A Re=Rp
B Rp=Rd
C Rd=Rα
D Rp=Rα

The correct answer is:
Rd=Rα

Solution:
In perpendicular magnetic field magnetic force provided centripetal force.
By the relation
qvB=mv2r
r=mvqB r∝mq
Since, mq ratio for deuteron and α-particle is the same.
mq=42=2 for α-particle

and (mq=21=2 for deuteron).
Hence, the radius is the same for both.

Topic: Physics, Magnetic Effects of Current, NEET
An electron accelerated through 500 V, enters a transverse uniform magnetic field of magnitude 100 mT. The radius of the circular path described by the electron is nearly

Options:
A 7.54Ã10-1 m
B 7.54×10-2 m
C 7.54×10-3 m
D 7.54×10-4 m

The correct answer is:
7.54×10-4 m

Solution:
r=2meVeB
=2meVace2B2
=2Ã9Ã10-31Ã5001.6Ã10-19Ã100Ã10-32
=2Ã9Ã10-31Ã5001.6Ã10-19Ã10-2
=2Ã9.1Ã5Ã10-31+231.6
=9.11.6Ã10-7
=9116Ã10-7
=91016Ã10-8
=56.875Ã10-4
=7.54Ã10-4 m

Topic: Physics, Magnetic Effects of Current, NEET
An electron accelerated under a potential difference of V volt has a certain wavelength Î». It is known that the mass of a proton is about 1800 times the mass of an electron. If a proton has to have the same wavelength Î», then it will have to be accelerated under the potential difference of

Options:
A V volt
B 1800Â volt
C V1800volt
D 1800Â volt

The correct answer is:
V1800volt

Solution:
Wavelength Î»=h2mqV
â mqV=h22Î»2
â mV=h22Î»2q=Constant
(since, Î» and q are same for electron and proton both)
â´ Vâ1m
â VPVe=memP
â VP=memP.Ve=V1800 volt

Topic: Physics, Atomic Physics, NEET
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).

Options:
A 12.65
B 13.54
C 14.55
D 17.97

The correct answer is:
13.54

Solution:
Correct Option is : (B)
13.54

Topic: Chemistry, Work Power Energy, NEET
An electron and a proton are situated in a uniform electric field. The ratio of their acceleration will be

Options:
A Zero
B Unity
C Ratio of masses of proton and electron
D Ratio of masses of electron and proton

The correct answer is:
Ratio of masses of proton and electron

Solution:
a=Fmforce exprienced by charge q in external electric field E is F=qEaeam=FemeFpmp=mp eEmeeE=mpme .

Topic: Physics, Electrostatics, NEET
An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicular to B, then

Options:
A Both will follow a circular path with same radius
B Both will follow a helical path
C Both will follow a parabolic path
D All the statements are false

The correct answer is:
All the statements are false

Solution:
For such a motion, magnetic force = centripetal force
qvB=mv2r
â qB=mvr
â r=mvqB
â´ rprc=mpvpqB ÃqBmeve
but Ke=Kp (given)
â 12 meve2=12mpvp2
â mpme=vevp2
â´ rpre=vevp2vpve=vevp
Hence, the electron and proton will follow circular paths but with different radii.

Topic: Physics, Magnetic Effects of Current, NEET
An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_{\varepsilon}$ and $\lambda_a$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

Options:
A $\lambda_e\gt\lambda_\alpha$
B $\lambda_e=4 \lambda_\alpha$
C $\lambda_e=\lambda \alpha$
D $\lambda_e \lt \lambda_a$

The correct answer is:
$\lambda_e\gt\lambda_\alpha$

Solution:
de-Broglie wavelength is given by
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}$
For same potential difference
$\lambda \propto \frac{1}{\sqrt{m q}}$
$\frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}$
$\because m_\alpha \gg m_e$
$\lambda_e\gt\lambda_\alpha$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2024 (Re-NEET)
An electron collides with a hydrogen atom in its ground state and excites it to n=3. The energy given to hydrogen atom in this inelastic collision is(neglect the motion of hydrogen atom)

Options:
A 10.2 eV
B 12.1 eV
C 12.5 eV
D None of these

The correct answer is:
12.1 eV

Solution:
The energy taken by hydrogen atom corresponds to its transition from
n= 1 to n=3 state.
∆E (given to hydrogen atom)
= 13.61-19
= 13.6×89 = 12.1 eV

Topic: Physics, Atomic Physics, NEET
An electron entering field normally with a velocity 4Ã107 m s-1 travels a distance of 0.10 m in an electric field of intensity 3200 V m-1. What is the deviation from its path?

Options:
A 1.76 mm
B 17.6Â mm
C 176 mm
D 0.176Â mm

The correct answer is:
1.76 mm

Solution:
Time to cross 0.10 m, t= 0.10 4× 10 7 s
Distance deviated in this time due to electric field = 1 2 ( eE m ).  t 2
= 1 2 × 1.6× 10 −19 ×3200 9.1× 10 −31 × ( 0.1 4× 10 7 ) 2 m
=01.76 mm

Topic: Physics, Electrostatics, NEET
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric filed is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

Options:
A 10 times greater
B 5 times greater
C Smaller
D Equal

The correct answer is:
Smaller

Solution:
Acceleration of charge q of mass m in uniform electric field is given as a=qEm.
Applying second equation of motion, S=ut+12at2.

Initially mass is at rest, so u=0.
h=0+12 qEmt2⇒t=2hmqE⇒t∝m
as mass of proton is heavier than electron, hence electron will take less time.

Topic: Physics, Electrostatics, NEET, NEET 2018
An electron flies into a homogeneous magnetic field of 10-3 T perpendicular to the force lines. The velocity of the electron is v=4Ã107 m s-1. What is the tangential acceleration of electron in the magnetic field?

Options:
A 7Ã1015 m s-2
B 7×1013 m s-2
C 7Ã1014 m s-2
D zero

The correct answer is:
zero

Solution:
Tangential acceleration is rate of change of speed. Work done by magnetic field on moving charge is zero. Therefore magnitude of velocity will not change and hence the tangential acceleration is zero.

Topic: Physics, Magnetic Effects of Current, NEET
An electron having kinetic energy 10 eV is circulating in a path of radius 0.1 m in an external magnetic field of intensity 10-4 T. The speed of the electron will be

Options:
A 1.76Ã106 ms-1
B 4.8Ã106Â ms-1
C 2.0Ã1012Â ms-1
D 4.8Ã1012 ms-1

The correct answer is:
1.76Ã106 ms-1

Solution:
When charge enters into a perpendicular magnetic field, it starts to move in a circular path.
Radius of a circular path followed by a charge particle:
r= mV qB
V=1.76Ã106 ms-1

Topic: Physics, Magnetic Effects of Current, NEET
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, NEET
An electron in C5+ ion during the transition from n=3 to n=1 emits light of wavelength

Options:
A 2.85 nm.
B 3.6 nm.
C 8.7 nm.
D 9.8 nm.

The correct answer is:
2.85 nm.

Solution:
Using Rydberg's equation for the transition of the electron,

1λ=RH Z2 1n12-1n22

Where,
RH=1.097×107 m-1

Z=6 for carbon

n1=1, n2=3

1λ=1.097×107×(6)2112-132 m-1

1λ=1.097×107×369-19 m-1

1λ=1.097×107×36×89 m-11λ=35.104×107 m-1

λ=135.104×107 mλ=0.0285×10-7 mλ=2.85×10-9 mλ=2.85 nm

Topic: Chemistry, Structure of Atom, NEET
An electron in hydrogen atom makes a transition $n_1 \rightarrow n_2$ where $n_1$ are principle quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1$ and $n_2$ are:

Options:
A $n_1=6$ and $n_2=2$
B $n_1=8$ and $n_2=1$
C $n_1=4$ and $n_2=2$
D $n_1=4$ and $n_2=2$

The correct answer is:
$n_1=4$ and $n_2=2$

Solution:
Time period of an electron in Bohr orbit is
$\begin{aligned}
& T=\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \\
\Rightarrow T & \propto n^3 \\
\Rightarrow \frac{T_1}{T_2} & =\frac{n_1^3}{n_2^3}
\end{aligned}$
As $T_1=8 T_2$ then
$\begin{aligned}
\Rightarrow & \left(\frac{8 T_2}{T_2}\right) & =\frac{n_1^3}{n_2^3} \\
\Rightarrow & \left(\frac{n_1}{n_2}\right)^3 & =\left(2^2\right)^3 \\
\Rightarrow & \frac{n_1}{n_2} & =2^2 \\
\Rightarrow & n_1 & =2 n_2
\end{aligned}$

Topic: Physics, Atomic Physics, NEET, NEET 2013 (Karnataka)
An electron in the H atom makes a transition from n=3 to n=1. The recoil momentum of the H atom will be

Options:
A 6.45Ã10-27 N s
B 6.8Ã10-27Â NÂ s
C 6.45Ã10-24Â NÂ s
D 6.8Ã10-24Â NÂ s

The correct answer is:
6.45Ã10-27 N s

Solution:
Energy of photon emitted
E=13.612-13.632
E=12.09 eV
Momentum of photon,
Ec=12.09Ã1.6Ã10-193Ã108 Nsâ¡=6.45Ã10-27 N s
Same will be the magnitude of recoil momentum of H atom

Topic: Physics, Atomic Physics, NEET
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10–2 nm, the potential difference is:

Options:
A 102 V
B 103 V
C 104 V
D 10 V

The correct answer is:
104 V

Solution:
λ=1.227×10−2 nm

=0.1227Å

λ=12.27vÅ

0.227=12.27vÅ

v=102⇒v=104 volt

Topic: Physics, Dual Nature of Matter, NEET, NEET 2020 (Phase 1)
An electron is accelerated through a potential difference of 45.5 V. The velocity acquired by it is (in m s-1â¡)

Options:
A 4×106
B 4×104
C 106
D zero

The correct answer is:
4×106

Solution:
If an electron is accelerated through a potential difference of V then its velocity is
v=2eVm
v=2Ã1.6Ã10-19Ã45.59.1Ã10-31
v=4Ã106 m s-1

Topic: Physics, Electrostatics, NEET
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly): me=9×10-31 kg

Options:
A 12.2×10-13 m
B 12.2×10-12 m
C 12.2×10-14 m
D 12.2 nm

The correct answer is:
12.2×10-12 m

Solution:
The energy of electron K=104 eV
λ=hmv=h2mK=6.63×10-342×9.1×10-31×104×1.6×10-19≅12.2×10-12 m

Topic: Physics, Dual Nature of Matter, NEET, NEET 2019 (All India)
An electron is accelerated under a potential difference of 64âV, the de-Broglie wavelength associated with electron is [e=-1.6Ã10â19 C, me=9.1Ã10â31 kg, h=6.623Ã10â34 Js]

Options:
A 1.53 A o
B 2.53 A o
C 3.35 A o
D 4.54 A o

The correct answer is:
1.53 A o

Solution:
Modified equation of de-Broglie wavelength Î»= h2meV
λ= 12.27 A o V = 12.27 64 =1.534  A o

Topic: Physics, Dual Nature of Matter, NEET
An electron is at ground state of the H atom. The minimum energy required to excite the H atom into the second excited state is

Options:
A 13.6  eV
B 12.1  eV
C 10.2  eV
D 3.4  eV

The correct answer is:
12.1  eV

Solution:
According to question, for an electron to jump from first excited state to second, He
energy required is, ÎE=E3-E1
=-13.632-(-13.6)
[âµ Energy of excited 2nd excited state, E3=-13.69]
â ÎE=-1.5+13.6=12.1 eV

Topic: Physics, Atomic Physics, NEET
An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV . The kinetic energy is E and its de Broglie wavelength is Î» . Then

Options:
A E = 6.8 eV, λ = 6.6 × 1 0 - 1 0 m
B E = 3.4 eV, Î» = 6.6 Ã 1 0 - 1 0 m
C E = 3.4 eV, λ = 6.6 × 1 0 - 1 1 m
D E = 6.8 eV, λ = 6.6 × 1 0 - 1 1 m

The correct answer is:
E = 3.4 eV, Î» = 6.6 Ã 1 0 - 1 0 m

Solution:
K.E.= KZ e 2 2r
P.E.= âKZ e 2 r
T.E.=P.E.+K.E.= âKZ e 2 2R
Therefore, TE=âKE= PE 2 =â3.4 eV
So, KE=3.14 eV
Let p = momentum and m = mass of the electron.
â´E= p 2 2m or p= 2mE
de Broglie wavelength,
Î»= h p = h 2mE
On substituting the values, we get
Î»= 6.63Ã 10 â34 2Ã9.1Ã 10 â31 Ã3.4Ã1.6Ã 10 â19
=6.6Ã 10 â10 m

Topic: Physics, Atomic Physics, NEET
An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is Î»

Options:
A E=6.8 eV, λ~6.6×10-10 m
B E=3.4 eV, Î»~6.6Ã10-10 m
C E=3.4 eV, Î»~6.6Ã10-11 m
D E=6.8 eV, Î»~6.6Ã10-11 m

The correct answer is:
E=3.4 eV, Î»~6.6Ã10-10 m

Solution:
PE=-2 KE=-2E
Total Energy = PE+KE=-2E+E =-E
â-E= -3.4 eV
or E=3.4 eV
Let p is momentum and m is mass of the electron
E=p22m
or p= (2mE) .
de Broglie wavelength Î»=hp=h(2mE)
Î»=6.6 x 10-10 m

Topic: Physics, Atomic Physics, NEET
An electron is in one of the 3d orbitals.Which of the following are possible values of n, l and mlfor this electron?

Options:
A n = 3 ; l = 3 ; ml= –3,–2, –1, 0, +1, +2, +3 (any one value)
B n = 3 ; l = 0 ; ml= 0 (any one value)
C n = 3 ; l = 1 ; ml= –1, 0, +1(any one value)
D n = 3 ; l = 2 ; ml= –2, –1, 0, +1, +2 (any one value)

The correct answer is:
n = 3 ; l = 2 ; ml= –2, –1, 0, +1, +2 (any one value)

Solution:
Correct Option is : (D)
n = 3 ; l = 2 ; ml= –2, –1, 0, +1, +2 (any one value)

Topic: Chemistry, Structure of Atom, NEET
An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The coulomb force $\vec{F}$ between the two is:
$\left(\right.$ where $\left.K=\frac{1}{4 \pi \varepsilon_0}\right)$

Options:
A $K \frac{e^2}{r^2} \hat{r}$
B $-K \frac{e^2}{r^3} \hat{r}$
C $K \frac{e^2}{r^3} \hat{r}$
D $-K \frac{e^2}{r^3} \hat{r}$

The correct answer is:
$-K \frac{e^2}{r^3} \hat{r}$

Solution:
The charge on hydrogen nuclear $q_1$
$\begin{aligned}
& =+v e \\
& \therefore \text { Coulomb's force } F=K \frac{q_1 q_2}{r^2} \\
& =K \frac{(+e)(-e)}{r^2}=-\frac{k e^2}{r^2} \hat{r}
\end{aligned}$

Topic: Physics, Electrostatics, NEET, NEET 2003
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57×10-2 T. If, the value of em is 1.76×1011 C kg-1, the frequency of revolution of the electron is

Options:
A 1  GHz
B 100  MHz
C 62.8  MHz
D 6.28  MHz

The correct answer is:
1  GHz

Solution:
f=qB2πm=1.76×1011×3.57×10-22×3.14

=109 Hz=1 GHz.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2016 (Phase 2)
An electron is moving with a kinetic energy of 4.55 Ã10-25 J. The de Broglie wavelength for it is
(Given: h=6.626× 10 −34 ,  m e =9.1× 10 −31  kg )

Options:
A 7.2×10-7m
B 72×10-7m
C 0.72×10-7m
D 4.2×10-7m

The correct answer is:
7.2×10-7m

Solution:
K.E.= 1 2   mv 2
V=2KEm
=2Ã4.55Ã10-259.1Ã10-31kg
= 10 3   mS −1
Î»=hmv=6.626Ã10-349.1Ã10-31Ã103
=7.2× 10 −7  m

Topic: Chemistry, Structure of Atom, NEET
An electron (mass = 9.1 Ã 10-31 ; 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, NEET