Atomic Structure – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions
Q. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ($\mathrm{Ch}=6.6 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$, mass of electron, $\mathrm{e}_{\mathrm{m}}=9.1 \times 10^{-31} \mathrm{kg}$):- (1) $1.92 \times 10^{-3} \mathrm{m}$ (2) $3.84 \times 10^{-3} \mathrm{m}$ (3) $1.52 \times 10^{-4} \mathrm{m}$ (4) $5.10 \times 10^{-3} \mathrm{m}$ [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) Q. Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^{3} \mathrm{ms}^{-1}$ (Mass of proton = $1.67 \times 10^{-27} \mathrm{kg}$ and $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$) :- (1) 2.5 nm (2) 14.0 nm (3) 0.032 nm (4) 0.40 nm [AIEEE-2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) $\mathrm{m}_{\mathrm{p}}=1.67 \times 10^{-27}$ $\mathrm{h}=6.63 \times 10^{-34}$ $\mathrm{v}=10^{3}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^{3}}$ $=3.97 \times 10^{-7+3}$ $=3.97 \times 10^{-10}$ $=\frac{3.9 \times 10^{-10}}{10^{-9}} \mathrm{nm} \quad=0.40 \mathrm{nm}$

Q. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ $\mathrm{mol}^{-1}$. The longest wavelength of light capable of breaking a single Cl–Cl bond is $\left(\mathrm{C}=3 \times 10^{8} \mathrm{ms}^{-1} \text { and } \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{mol}^{-1}\right)$ (1) 494 nm (2) 594 nm (3) 640 nm (4) 700 nm [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{B.E.}=242 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{E}=\frac{\mathrm{hcN}_{\mathrm{A}}}{\lambda}$ $10^{3} \times 242 \times \lambda=3 \times 10^{8} \times 6.626 \times 10^{-34} \times 6.02 \times 10^{23}$ $\lambda=\frac{3 \times 6.626 \times 6.02 \times 10^{-26+23}}{242}$ $=0.494 \times 10^{-3} \times 10^{-3}$ = 494 nm

Q. Ionisation energy of $\mathrm{He}^{+}$ is $19.6 \times 10^{-18} \mathrm{J}$ atom $^{-1}$. The energy of the first stationary state (n = 1) of $\mathrm{L} \mathbf{i}^{2+}$ is:- (1) $8.82 \times 10^{-17} \mathrm{J}$ atom $^{-1}$ (2) $4.41 \times 10^{-16} \mathrm{J}$ atom $^{-1}$ (3) $-4.41 \times 10^{-17} \mathrm{J}$ atom $^{-1}$ (4) $-2.2 \times 10^{-15} \mathrm{J}$ atom $^{-1}$ [AIEEE-2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) I.E. $=19.6 \times 10^{-18}$ I.E $\propto \mathrm{z}^{2}$ $\frac{(\mathrm{I.E.})_{\mathrm{Li}^{+2}}}{(\mathrm{I.E.})_{\mathrm{He}}}=\frac{\mathrm{Z}_{\mathrm{Li}}^{2}}{\mathrm{Z}_{\mathrm{He}}^{2}} \quad \mathrm{E}_{1}=\frac{9}{4} \times 19.6 \times 10^{-18}$ $=-4.41 \times 10^{-17}$

Q. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following (1) n = 3 to n = 1 (2) n = 2 to n = 1 (3) n = 3 to n = 2 (4) n = 4 to n = 3 [AIEEE-2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2)  Q. The electrons identified by quantum numbers n and  :- (a) n = 4 ,  = 1 (b) n = 4,  = 0 (c) n = 3,  = 2 (d) n = 3,  = 1 Can be placed in order of increasing energy as (1) (a) < (c) < (b) < (d) (2) (c) < (d) < (b) < (a) (3) (d) < (b) < (c) < (a) (4) (b) < (d) < (a) < (c) (3) (d) < (b) < (c) < (a) (4) (b) < (d) < (a) < (c) [AIEEE-2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) (d) < (b) < (c) < (a) Acc. to (n + ) rule.

Q. If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become :- (1) Two times (2) Half (3) One fourth (4) Four time [JEE-Main(online2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}}$

Q. The wave number of the first emission line in the Balmer series of H-Spectrum is : (R = Rydberg constant) : (1) $\frac{3}{4} \mathrm{R}$ (2) $\frac{9}{400} \mathrm{R}$ (3) $\frac{5}{36} \mathrm{R}$ (4) $\frac{7}{6} \mathrm{R}$ [JEE-Main(online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\bar{v}=\frac{1}{\mathrm{R}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5}{36 \mathrm{R}}$

Q. The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is : $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{J} \mathrm{s}\right)$ (1) $6.626 \times 10^{-31} \mathrm{m}$ (2) $6.626 \times 10^{-34} \mathrm{m}$ (3) $6.626 \times 10^{-38} \mathrm{m}$ (4) $6.626 \times 10^{-30} \mathrm{m}$ [JEE-Main(online) 2013]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)

Q. For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship? (1) a dust particle (2) an electron (3) a proton (4) an -particle. [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1)

Q. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of $\mathbf{L} \mathbf{i}^{++}$ is : (1) 13.6 eV (2) 30.6 eV (3) 122.4 eV (4) 3.4 eV [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) B.E. $=3.4 \times 9=30.6 \mathrm{eV}$

Q. Based on the equation $\Delta \mathrm{E}=-2.0 \times 10^{-18} \mathrm{J}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n $=2$ will be $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}, \mathrm{C}=3 \times 10^{8} \mathrm{ms}^{-1}\right)$ (1) $2.650 \times 10^{-7} \mathrm{m}$ (2) $1.325 \times 10^{-7} \mathrm{m}$ (3) $1.325 \times 10^{-10} \mathrm{m}$ (4) $5.300 \times 10^{-10} \mathrm{m}$ [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (2) $\frac{1}{\lambda}=\frac{2 \times 10^{-18}}{\mathrm{hc}}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\Rightarrow \frac{1}{\lambda}=\frac{2 \times 10^{-18}}{6.625 \times 10^{-34} \times 3 \times 10^{8}} \times \frac{3}{4}$ $\Rightarrow \lambda=\frac{2 \times 6.625 \times 10^{-34} \times 10^{8}}{10^{-18}}$ $=13.25 \times 10^{-8}$ $=1.325 \times 10^{-7} \mathrm{m}$

Q. If be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) $\mathrm{E}=\phi+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \mathrm{v}^{2}=\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right] \Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]}$ $\Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]}$

Q. Ionization energy of gaseous Na atoms is 495.5 $\mathrm{kjmol}^{-1}$ . The lowest possible frequency of light that ionizes a sodium atom is $\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$ (1) $3.15 \times 10^{15} \mathrm{s}^{-1}$ (2) $4.76 \times 10^{14} \mathrm{s}^{-1}$ (3) $1.24 \times 10^{15} \mathrm{s}^{-1}$ (4) $7.50 \times 10^{4} \mathrm{s}^{-1}$ [JEE-Main(online) 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3) $\Delta \mathrm{E}=\mathrm{hv}$ $\mathrm{v}=\frac{\Delta \mathrm{E}}{\mathrm{h}}$ $\mathrm{v}=\frac{495.5 \times 10^{3} \mathrm{Joule}}{6.023 \times 10^{23}} \times \frac{1}{6.626 \times 10^{-34}}$ $\mathrm{v}=1.24 \times 10^{15} \mathrm{sec}^{-1}$

Q. Which of the following is the energy of a possible excited state of hydrogen? (1) –3.4 eV (2) +6.8 eV (3) +13.6 eV (4) –6.8 eV [JEE-Main(offline) 2015]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) For H-atom, (Z = 1) $\mathrm{E}_{\mathrm{n}}=-13.6 \times \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$ \begin{aligned} \therefore \text { So for } \mathrm{E}_{1} &=-13.6 \mathrm{eV} \\ \mathrm{E}_{2} &=-3.4 \mathrm{eV} \end{aligned}

Q. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron respectively, then the value of $\mathrm{h} / \lambda$ (where $\lambda$is wavelength associated with electron wave) is given by : (1) $\sqrt{2 \mathrm{meV}}$ (2) mev (3) $2 \mathrm{meV}$ (4) $\sqrt{\mathrm{meV}}$ [JEE-Main 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (1) As electron of charge ‘e’ is passed through ‘V’ volt, kinetic energy of electron becomes = ‘eV’ As wavelength of $e^{-}$ wave = $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m} \cdot \mathrm{K} \cdot \mathrm{E}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\therefore \quad \frac{\mathrm{h}}{\lambda}=\sqrt{2 \mathrm{meV}}$

Q. The radius of the second Bohr orbit for hydrogen atom is : (Planks const. $\mathrm{h}=6.6262 \times 10^{-34} \mathrm{Js}$; mass of electron $=9.1091 \times 10^{-31} \mathrm{kg} ;$ charge of electron $\mathrm{e}=$ $1.60210 \times 10^{-19} \mathrm{C}:$ permittivity of vaccuml $\left.\epsilon_{0}=8.854185 \times 10^{-12} \mathrm{kg}^{-1} \mathrm{m}^{-3} \mathrm{A}^{2}\right)$ [JEE-Main 2017]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4) Radius of $\mathbf{n}^{\mathrm{th}}$ Bohr orbit in H-atom Radius of II Bohr orbit $=0.53 \times(2)^{2}$ • October 1, 2021 at 9:09 pm

Thanks for this

1
• June 26, 2021 at 6:46 am

Op

14
• June 15, 2021 at 1:45 pm

prompt(1)

1
• June 15, 2021 at 1:44 pm

this is bold

5
• October 17, 2021 at 11:51 am

Good

0
• May 31, 2021 at 5:49 am

Very useful information regarding to atomic structure I prefer esaral for this kind of solutions and questions

3
• May 26, 2021 at 10:43 am

There are good but less in number.

80
• May 26, 2021 at 10:45 am
• April 24, 2021 at 9:12 pm

Excellent questions

53
• February 28, 2021 at 8:08 pm

Nice really I had nice revision thank u very much esaral

102
• February 28, 2021 at 8:09 pm

Yeah really

1
• February 16, 2021 at 3:58 pm

I LIKE THIS WEBSITE BCZ THEY PROVIDE QUESTIONS SPECIFICALLY WHICH WERE ASKED IN JEE ONLY SO DURING EXAM TIME WE CAN KNOW THAT WHAT KING OF QUESTION WILL BE GONNA ASK IN JEE

49
• January 6, 2021 at 10:08 pm

nice questions

10
• November 18, 2020 at 7:48 pm

good questions but we need some m ore practice

5
• November 19, 2020 at 10:50 am

For Complete study material, Video Lectures and Mind Map Download eSaral APP: https://esaral.com/app .

12
• October 28, 2020 at 8:11 am

Nice question

3
• October 11, 2020 at 11:53 am

good but we need some more questions

2
• September 15, 2020 at 12:14 am

Thanks

70
• September 13, 2020 at 1:07 am

thnx bro!
appreciated

70
• September 9, 2020 at 6:47 pm

Thank u man……

31
• September 8, 2020 at 10:46 pm

It is very helpful for me👍👍👍

1
• September 6, 2020 at 2:13 pm

0
• December 8, 2020 at 7:24 pm

Love u guys for amazing content

0
• September 4, 2020 at 10:48 am

thank you so much

7
• September 2, 2020 at 5:30 pm

🤘😝🤘Fantastic questions

4
• August 28, 2020 at 1:11 pm

Niceeeeeeeeee!!!!!!!!!!!
THANK YOU!

69
• August 27, 2020 at 11:01 pm

Nice questions👍👍

69
• August 24, 2020 at 4:41 pm

Nice questions. Thank you for your help.

0
• August 22, 2020 at 11:52 pm

0
• August 19, 2020 at 10:52 am

Hi thanks you ☺️🙂😽

0
• August 11, 2020 at 7:48 pm

We owe a large time man…!

0
• July 23, 2020 at 3:43 pm

it was awesome sir

0
• July 10, 2020 at 11:06 am

👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻

0
• July 7, 2020 at 2:25 pm

THANK YOU IT WAS VERY HELPFULL

0
• July 1, 2020 at 8:34 pm

It was very useful🙂🙂

0
• June 30, 2020 at 12:43 am

0
• June 27, 2020 at 7:42 am

Nice

0
• June 15, 2020 at 7:42 pm

Very interesting sets of questions

0
• May 29, 2020 at 4:28 pm

I good practice on this chapter so thank you soon muchhhh

0
• May 26, 2020 at 7:52 am

very nice

0
• May 24, 2020 at 2:06 pm

Chill bro

0
• May 22, 2020 at 8:45 pm

it was really helpfull…. thanks a lot

0
• May 21, 2020 at 9:53 pm

Thanks

4
• May 19, 2020 at 8:35 pm

Thank You….It was very useful

0
• May 17, 2020 at 7:13 am

Thank you

0
• May 7, 2020 at 11:36 pm

Maa Chudao

0
• May 4, 2020 at 3:45 pm

Goooooooooooooooooooooooooood

0
• April 28, 2020 at 6:40 pm

Ok

0
• April 27, 2020 at 9:06 pm

vrry goooddd

0
• April 16, 2020 at 2:59 pm

Thanks

0
• April 7, 2020 at 1:03 pm

It’s nice

0
• April 6, 2020 at 10:34 am

Thanks

0
• April 4, 2020 at 10:53 pm

Thank you

0
• March 18, 2020 at 8:48 pm

awesome questions

0
• November 1, 2019 at 6:04 pm

super

0