Atomic Structure - JEE Main Previous Year Questions with Solutions

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JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.   Simulator   Previous Years AIEEE/JEE Mains Questions [esquestion] 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) $5.10 \times 10^{-3} \mathrm{m}$ #tag# [AIEEE-2009] #sol# (1) [/esquestion] [esquestion] 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 #tag# [AIEEE-2009] #sol# (4) $\mathrm{m}_{\mathrm{p}}=1.67 \times 10^{-27}$ $\mathrm{h}=6.63 \times 10^{-34}$ $\mathrm{v}=10^{3}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^{3}}$ $=3.97 \times 10^{-7+3}$ $=3.97 \times 10^{-10}$ $=\frac{3.9 \times 10^{-10}}{10^{-9}} \mathrm{nm} \quad=0.40 \mathrm{nm}$ [/esquestion] [esquestion] 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 #tag# [AIEEE-2010] #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 [/esquestion] [esquestion] 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 $^{-1}$ #tag# [AIEEE-2010] #sol# (3) I.E. $=19.6 \times 10^{-18}$ I.E $\propto \mathrm{z}^{2}$ $\frac{(\mathrm{I.E.})_{\mathrm{Li}^{+2}}}{(\mathrm{I.E.})_{\mathrm{He}}}=\frac{\mathrm{Z}_{\mathrm{Li}}^{2}}{\mathrm{Z}_{\mathrm{He}}^{2}} \quad \mathrm{E}_{1}=\frac{9}{4} \times 19.6 \times 10^{-18}$ $=-4.41 \times 10^{-17}$ [/esquestion] [esquestion] 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 #tag# [AIEEE-2011] #sol# (2) [/esquestion] [esquestion] 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) #tag# [AIEEE-2012] #sol# (3) (d) < (b) < (c) < (a) Acc. to (n + ) rule. [/esquestion] [esquestion] 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 #tag# [JEE-Main(online2012] #sol# (2) $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}}$ [/esquestion] [esquestion] 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) $\frac{7}{6} \mathrm{R}$ #tag# [JEE-Main(online) 2013] #sol# (3) $\bar{v}=\frac{1}{\mathrm{R}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5}{36 \mathrm{R}}$ [/esquestion] [esquestion] 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) $6.626 \times 10^{-30} \mathrm{m}$ #tag# [JEE-Main(online) 2013] #sol# (3) [/esquestion] [esquestion] 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. #tag# [JEE-Main(online) 2014] #sol# (1) [/esquestion] [esquestion] 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 #tag# [JEE-Main(online) 2014] #sol# (2) B.E. $=3.4 \times 9=30.6 \mathrm{eV}$ [/esquestion] [esquestion] 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) $5.300 \times 10^{-10} \mathrm{m}$ #tag# [JEE-Main(online) 2014] #sol# (2) $\frac{1}{\lambda}=\frac{2 \times 10^{-18}}{\mathrm{hc}}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\Rightarrow \frac{1}{\lambda}=\frac{2 \times 10^{-18}}{6.625 \times 10^{-34} \times 3 \times 10^{8}} \times \frac{3}{4}$ $\Rightarrow \lambda=\frac{2 \times 6.625 \times 10^{-34} \times 10^{8}}{10^{-18}}$ $=13.25 \times 10^{-8}$ $=1.325 \times 10^{-7} \mathrm{m}$ [/esquestion] [esquestion] If be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is #tag# [JEE-Main(online) 2014] #sol# (1) $\mathrm{E}=\phi+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{1}{2} \mathrm{mv}^{2}$ $\Rightarrow \mathrm{v}^{2}=\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right] \Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]}$ $\Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}}\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]}$ [/esquestion] [esquestion] 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) $7.50 \times 10^{4} \mathrm{s}^{-1}$ #tag# [JEE-Main(online) 2014] #sol# (3) $\Delta \mathrm{E}=\mathrm{hv}$ $\mathrm{v}=\frac{\Delta \mathrm{E}}{\mathrm{h}}$ $\mathrm{v}=\frac{495.5 \times 10^{3} \mathrm{Joule}}{6.023 \times 10^{23}} \times \frac{1}{6.626 \times 10^{-34}}$ $\mathrm{v}=1.24 \times 10^{15} \mathrm{sec}^{-1}$ [/esquestion] [esquestion] 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 #tag# [JEE-Main(offline) 2015] #sol# (1) For H-atom, (Z = 1) $\mathrm{E}_{\mathrm{n}}=-13.6 \times \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}$ $\begin{aligned} \therefore \text { So for } \mathrm{E}_{1} &=-13.6 \mathrm{eV} \\ \mathrm{E}_{2} &=-3.4 \mathrm{eV} \end{aligned}$ [/esquestion] [esquestion] 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) $\sqrt{\mathrm{meV}}$ #tag# [JEE-Main 2016] #sol# (1) As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes = 'eV' As wavelength of $e^{-}$ wave = $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m} \cdot \mathrm{K} \cdot \mathrm{E}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\therefore \quad \frac{\mathrm{h}}{\lambda}=\sqrt{2 \mathrm{meV}}$ [/esquestion] [esquestion] 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 $\left.\epsilon_{0}=8.854185 \times 10^{-12} \mathrm{kg}^{-1} \mathrm{m}^{-3} \mathrm{A}^{2}\right)$ #tag# [JEE-Main 2017] #sol# (4) Radius of $\mathbf{n}^{\mathrm{th}}$ Bohr orbit in H-atom Radius of II Bohr orbit $=0.53 \times(2)^{2}$ [/esquestion]

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