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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is
[JEE-Main(online) 2014]
Ans. (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]
Ans. (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]
Ans. (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]
Ans. (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]
[JEE-Main 2017]
Ans. (4)
Radius of $\mathbf{n}^{\mathrm{th}}$ Bohr orbit in H-atom
Radius of II Bohr orbit $=0.53 \times(2)^{2}$
Radius of II Bohr orbit $=0.53 \times(2)^{2}$
Comments
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May 10, 2024, 6:35 a.m.
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..
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May 31, 2021, 5:49 a.m.
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VAIDIK
Feb. 16, 2021, 3:58 p.m.
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