{"id":866208,"date":"2025-08-04T15:49:29","date_gmt":"2025-08-04T10:19:29","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866208"},"modified":"2025-08-04T15:49:29","modified_gmt":"2025-08-04T10:19:29","slug":"ncert-solutions-class-11-chemistry-part-i-chapter-2-structure-of-atom-free-pdf","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-i-chapter-2-structure-of-atom-free-pdf\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-I) Chapter-2: Structure of Atom (Free PDF)"},"content":{"rendered":"\n<p>In this unit, we have earlier provided you with a summary of notes on the structure of atoms that includes building blocks of matter, including their properties, various atomic theories, quantum theories of the atom, and the principles of electronic configuration. However, without practicing with unit-based questions, you may lack clarity and a deeper understanding. This blog provides some exercises along with their solutions to help you understand the concepts more easily.<\/p>\n\n\n\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#aadbf9\"><tbody><tr><td><strong>Chapter 1<\/strong><\/td><td><strong>Chapter 2<\/strong><\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-class-11-chemistry-part-i-chapter-2-structure-of-atom\">NCERT Solutions Class 11 Chemistry (Part-I) Chapter 2: Structure of Atom<\/h2>\n\n\n\n<p>Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part I) Chapter 2: Structure of Atom.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-exercises\">Exercises<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li>(i) Calculate the number of electrons that will weigh one gram together. (ii) Calculate the mass and charge of one mole of electrons.<\/li>\n\n\n\n<li>(i) Calculate the total number of electrons present in one mole of methane. (ii) Find<br>\u2003(a) The total number and<br>\u2003(b) The total mass of neutrons in 7 mg of \u00b9\u2074C.<br>(iii) Find<br>\u2003(a) The total number and<br>\u2003(b) The total mass of protons in 34 mg of NH\u2083 at STP.<\/li>\n\n\n\n<li>How many neutrons and protons are there in the following nuclei:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u2086\u00b9\u00b3C<\/li>\n\n\n\n<li>\u2088\u00b9\u2076O<\/li>\n\n\n\n<li>\u2081\u2082\u00b2\u2074Mg<\/li>\n\n\n\n<li>\u2082\u2086\u2075\u2076Fe<\/li>\n\n\n\n<li>\u2083\u2088\u2078\u2078Sr<\/li>\n<\/ul>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li>Will the answer change if the temperature and pressure are changed?<\/li>\n<\/ol>\n\n\n\n<p>Write the complete symbol for the atom with the given atomic number (Z) and mass number (A):<br>(i) Z = 17, A = 35<br>(ii) Z = 92, A = 233<br>(iii) Z = 4, A = 9<\/p>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li>Yellow light emitted from a sodium lamp has a wavelength (\u03bb) of 580 nm. Calculate the frequency (\u03bd) and wavenumber (\u03bd\u0305) of the yellow light.<\/li>\n\n\n\n<li>Find the energy of each of the photons which:<br>(i) Correspond to light of frequency 3 \u00d7 10\u00b9\u2075 Hz<br>(ii) Have a wavelength of 0.50 \u00c5<\/li>\n\n\n\n<li>Calculate the wavelength, frequency, and wavenumber of a light wave whose period is 2.0 \u00d7 10\u207b\u00b9\u2070 s.<\/li>\n\n\n\n<li>What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?<\/li>\n\n\n\n<li>A photon of wavelength 4 \u00d7 10\u207b\u2077 m strikes a metal surface. The work function of the metal is 2.13 eV. Calculate:<br>(i) Energy of the photon (eV)<br>(ii) Kinetic energy of the emission<br>(iii) Velocity of the photoelectron<br>(1 eV = 1.6020 \u00d7 10\u207b\u00b9\u2079 J)<\/li>\n\n\n\n<li>Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionization energy of sodium in kJ mol\u207b\u00b9.<\/li>\n\n\n\n<li>A 25-watt bulb emits monochromatic yellow light of wavelength 0.57 \u00b5m. Calculate the rate of emission of quanta per second.<\/li>\n\n\n\n<li>Electrons are emitted with zero velocity from a metal surface when exposed to radiation of wavelength 6800 \u00c5. Calculate:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Threshold frequency (\u03bd\u2080)<\/li>\n\n\n\n<li>Work function (W\u2080) of the metal.<\/li>\n<\/ul>\n\n\n\n<ol start=\"13\" class=\"wp-block-list\">\n<li>What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from n = 4 to n = 2?<\/li>\n\n\n\n<li>How much energy is required to ionise a hydrogen atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of a hydrogen atom from n = 1 orbit.<\/li>\n\n\n\n<li>What is the maximum number of emission lines when an excited electron of a hydrogen atom in n = 6 drops to the ground state?<\/li>\n\n\n\n<li>(i) The energy associated with the first orbit in a hydrogen atom is \u20132.18 \u00d7 10\u207b\u00b9\u2078 J atom\u207b\u00b9. What is the energy associated with the fifth orbit?<\/li>\n\n\n\n<li>Calculate the wavenumber for the longest wavelength transition in the Balmer series of hydrogen.<\/li>\n\n\n\n<li>What is the energy (in J) required to shift the electron of hydrogen from the first Bohr orbit to the fifth? Also, find the wavelength of light emitted when it returns to the ground state.<br>(Ground state energy = \u20132.18 \u00d7 10\u207b\u00b9\u00b9 ergs)<\/li>\n\n\n\n<li>Using E\u2099 = \u20132.18 \u00d7 10\u207b\u00b9\u2078 \/ n\u00b2 J, calculate the energy required to remove an electron from the n = 2 orbit. What is the longest wavelength of light (in cm) that can cause this transition?<\/li>\n\n\n\n<li>Calculate the wavelength of an electron moving with a velocity of 2.05 \u00d7 10\u2077 m\/s.<\/li>\n\n\n\n<li>The mass of an electron is 9.1 \u00d7 10\u207b\u00b3\u00b9 kg. If its kinetic energy is 3.0 \u00d7 10\u207b\u00b2\u2075 J, calculate its wavelength.<\/li>\n\n\n\n<li>(i) Which of the following are isoelectronic species?<br>Na\u207a, K\u207a, Mg\u00b2\u207a, Ca\u00b2\u207a, S\u00b2\u207b, Ar (ii). Calculate the radius of Bohr\u2019s fifth orbit for the hydrogen atom.<\/li>\n\n\n\n<li>(i) Write the electronic configurations of the following ions:<br>(a) H\u207b (b) Na\u207a (c) O\u00b2\u207b (d) F\u207b (ii) What are the atomic numbers of elements with outermost electrons represented by:<br>(a) 3s\u00b9 (b) 2p\u00b3 (c) 3p\u2075 (iii) Which atoms are indicated by these configurations?<br>(a) [He] 2s\u00b9<br>(b) [Ne] 3s\u00b2 3p\u00b3<br>(c) [Ar] 4s\u00b2 3d\u00b9<\/li>\n\n\n\n<li>What is the lowest value of n that allows g-orbitals to exist?<\/li>\n\n\n\n<li>An electron is in one of the 3d orbitals. Give the possible values of n, l and m\u2097 for this electron.<\/li>\n\n\n\n<li>An atom of an element contains 29 electrons and 35 neutrons. Deduce:<br>(i) Number of protons<br>(ii) Electronic configuration of the element<\/li>\n\n\n\n<li>(i) An atomic orbital has n = 3. What are the possible values of l and m\u2097?<br>(ii) List quantum numbers (l and m\u2097) of electrons in the 3d orbital.<br>(iii) Which of the following orbitals are possible: 1p, 2s, 2p, 3f?<\/li>\n\n\n\n<li>Using s, p, d notation, describe orbitals with the following quantum numbers:<br>(a) n = 1, l = 0<br>(b) n = 3, l = 1<br>(c) n = 4, l = 2<br>(d) n = 4, l = 3<\/li>\n\n\n\n<li>How many electrons in an atom may have the following quantum numbers?<br>(a) n = 4, m\u209b = \u2013\u00bd<br>(b) n = 3, l = 0<\/li>\n\n\n\n<li>Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.<\/li>\n\n\n\n<li>What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of the He\u207a spectrum?<\/li>\n\n\n\n<li>Calculate the energy required for the process:<br>He\u207a(g) \u2192 He\u00b2\u207a(g) + e\u207b<br>(Ionisation energy for H atom in the ground state is 2.18 \u00d7 10\u207b\u00b9\u2078 J atom\u207b\u00b9)<\/li>\n\n\n\n<li>If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms that can be placed side by side in a straight line across a 20 cm long scale.<\/li>\n\n\n\n<li>2 \u00d7 10\u2078 atoms of carbon are arranged side by side. Calculate the radius of a carbon atom if the length of this arrangement is 2.4 cm.<\/li>\n\n\n\n<li>The diameter of a zinc atom is 2.6 \u00c5. Calculate:<br>(a) Radius of the zinc atom in pm<br>(b) Number of atoms present in a length of 1.6 cm if zinc atoms are arranged side by side lengthwise<\/li>\n\n\n\n<li>A certain particle carries 2.5 \u00d7 10\u207b\u00b9\u2076 C of static electric charge. Calculate the number of electrons present in it.<\/li>\n\n\n\n<li>In Millikan\u2019s experiment, the static electric charge on an oil drop was \u20131.282 \u00d7 10\u207b\u00b9\u2078 C. Calculate the number of electrons present on it.<\/li>\n\n\n\n<li>In Rutherford\u2019s experiment, thin foils of heavy atoms like gold or platinum were bombarded with \u03b1-particles. What difference in results would be observed if a foil of light atoms like aluminum were used instead?<\/li>\n\n\n\n<li>Write the correct symbolic representation for:<br>79\/35 Br and explain why 79Br or 35Br alone are not acceptable.<\/li>\n\n\n\n<li>An element with mass number 81 contains 31.7% more neutrons than protons. Assign the atomic symbol.<\/li>\n\n\n\n<li>An ion with mass number 37 has one unit of negative charge and contains 11.1% more neutrons than electrons. Find the symbol of the ion.<\/li>\n\n\n\n<li>An ion with mass number 56 has a 3+ charge and contains 30.4% more neutrons than electrons. Assign the symbol to this ion.<\/li>\n\n\n\n<li>Arrange the following types of radiation in increasing order of frequency:<br>(a) Radiation from a microwave oven<br>(b) Amber light from the traffic signal<br>(c) Radiation from FM radio<br>(d) Cosmic rays from outer space<br>(e) X-rays<\/li>\n\n\n\n<li>A nitrogen laser produces radiation at a wavelength of 337.1 nm. If it emits 5.6 \u00d7 10\u00b2\u2074 photons, calculate the power of this laser.<\/li>\n\n\n\n<li>Neon gas is used in signboards. If it emits strongly at 616 nm, calculate:<br>(a) Frequency of emission<br>(b) Distance travelled by this radiation in 30 s<br>(c) Energy of one quantum<br>(d) Number of quanta if 2 J of energy is produced<\/li>\n\n\n\n<li>In astronomical observations, signals from distant stars are weak. If the photon detector receives 3.15 \u00d7 10\u207b\u00b9\u2078 J from 600 nm radiation, calculate the number of photons received.<\/li>\n\n\n\n<li>A pulsed radiation source of 2 ns emits 2.5 \u00d7 10\u00b9\u2075 photons. Calculate the energy of the source.<\/li>\n\n\n\n<li>The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Frequency of each transition<\/li>\n\n\n\n<li>Energy difference between the two excited states<\/li>\n<\/ul>\n\n\n\n<ol start=\"49\" class=\"wp-block-list\">\n<li>The work function for the cesium atom is 1.9 eV. Calculate:<br>(a) Threshold wavelength<br>(b) Threshold frequency<br>Also, if irradiated with 500 nm light, calculate the kinetic energy and velocity of the ejected photoelectron.<\/li>\n\n\n\n<li>Sodium is irradiated with light of wavelengths 500 nm, 450 nm, and 400 nm. Corresponding photoelectron velocities (\u00d710\u2075 cm\/s): 2.55, 4.35, 5.35.<br>Calculate:<br>(a) Threshold wavelength<br>(b) Planck\u2019s constant<\/li>\n\n\n\n<li>In a photoelectric experiment, silver ejects electrons stopped by 0.35 V when irradiated with 256.7 nm radiation. Calculate the work function of silver.<\/li>\n\n\n\n<li>A photon of 150 pm ejects an inner electron with velocity 1.5 \u00d7 10\u2077 m\/s. Calculate the binding energy of this electron to the nucleus.<\/li>\n\n\n\n<li>Paschen series transitions end at n = 3 and are given by:<br>\u03bd = 3.29 \u00d7 10\u00b9\u2075 Hz \u00d7 [1\/3\u00b2 \u2013 1\/n\u00b2]<br>Calculate the value of n for the 1285 nm transition and the region of the spectrum.<\/li>\n\n\n\n<li>A transition starts from an orbit of radius 1.3225 nm and ends at 211.6 pm.<br>Calculate:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Wavelength of emission<\/li>\n\n\n\n<li>Name of the series<\/li>\n\n\n\n<li>Region of the spectrum<\/li>\n<\/ul>\n\n\n\n<ol start=\"55\" class=\"wp-block-list\">\n<li>An electron microscope uses electrons at a velocity of 1.6 \u00d7 10\u2076 m\/s. Calculate the de Broglie wavelength of the electron.<\/li>\n\n\n\n<li>Neutron diffraction uses 800 pm wavelength. Calculate the characteristic velocity of the neutron.<\/li>\n\n\n\n<li>If the velocity of an electron in Bohr\u2019s first orbit is 2.19 \u00d7 10\u2076 m\/s, calculate the associated de Broglie wavelength.<\/li>\n\n\n\n<li>A proton moving at 4.37 \u00d7 10\u2075 m\/s (from 1000 V potential). A hockey ball of mass 0.1 kg moves at the same speed. Calculate the wavelength associated with this velocity for both.<\/li>\n\n\n\n<li>If the position of an electron is known within \u00b10.002 nm, calculate the uncertainty in momentum. Also, check consistency with momentum defined as h \/ (4\u03c0m \u00d7 0.05 nm).<\/li>\n\n\n\n<li>Given the quantum numbers of six electrons:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>n = 4, l = 2, m\u2097 = \u20132, m\u209b = \u2013\u00bd<\/li>\n\n\n\n<li>n = 3, l = 2, m\u2097 = 1, m\u209b = +\u00bd<\/li>\n\n\n\n<li>n = 4, l = 1, m\u2097 = 0, m\u209b = +\u00bd<\/li>\n\n\n\n<li>n = 3, l = 2, m\u2097 = \u20132, m\u209b = \u2013\u00bd<\/li>\n\n\n\n<li>n = 3, l = 1, m\u2097 = \u20131, m\u209b = +\u00bd<\/li>\n\n\n\n<li>n = 4, l = 1, m\u2097 = 0, m\u209b = +\u00bd<\/li>\n<\/ul>\n\n\n\n<p>Arrange in order of increasing energy and identify any degenerate states.<\/p>\n\n\n\n<ol start=\"61\" class=\"wp-block-list\">\n<li>Br (Z = 35) has:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>6 electrons in 2p<\/li>\n\n\n\n<li>6 in 3p<\/li>\n\n\n\n<li>5 in 4p<\/li>\n<\/ul>\n\n\n\n<p>Which electron experiences the lowest effective nuclear charge?<\/p>\n\n\n\n<ol start=\"62\" class=\"wp-block-list\">\n<li>Among the following pairs, which orbital experiences a larger effective nuclear charge?<br>(i) 2s and 3s<br>(ii) 4d and 4f<br>(iii) 3d and 3p<\/li>\n\n\n\n<li>Unpaired electrons in Al and Si are in 3p orbitals. Which experiences a more effective nuclear charge?<\/li>\n\n\n\n<li>Indicate the number of unpaired electrons in:<br>(a) P\u2003(b) Si\u2003(c) Cr\u2003(d) Fe\u2003(e) Kr<\/li>\n\n\n\n<li>(a) How many subshells are associated with n = 4?<br>(b) How many electrons will be present in the subshells having m\u209b = \u2013\u00bd for n = 4?<\/li>\n<\/ol>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-geography-fundamentals-of-physical-geography-chapter-12-water-oceans\/\"><strong>NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 12: Water (Oceans) (Free PDF)<\/strong><\/a><\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-solutions\">Solutions<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Mass of 1 electron = 9.109 \u00d7 10\u207b\u00b3\u00b9 kg<br>1 g = 10\u207b\u00b3 kg<br>Number of electrons = (10\u207b\u00b3) \/ (9.109 \u00d7 10\u207b\u00b3\u00b9) = 1.098 \u00d7 10\u00b2\u2077 electrons<\/li>\n\n\n\n<li>Mass of 1 mole of electrons = 9.109 \u00d7 10\u207b\u00b3\u00b9 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 5.485 \u00d7 10\u207b\u2074 kg<br>Charge of 1 mole of electrons = 1.602 \u00d7 10\u207b\u00b9\u2079 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 9.648 \u00d7 10\u2074 C<\/li>\n\n\n\n<li>Electrons in 1 CH\u2084 molecule = 6 (C) + 4 (H) = 10<br>1 mole of methane has 6.022 \u00d7 10\u00b2\u00b3 molecules<br>Total electrons = 10 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 6.022 \u00d7 10\u00b2\u2074 electrons<\/li>\n\n\n\n<li>Neutrons in one atom of \u00b9\u2074C = 14 \u2013 6 = 8<br>Mass of \u00b9\u2074C = 14 g\/mol<br>Moles in 7 mg = (7 \u00d7 10\u207b\u00b3) \/ 14 = 0.5 \u00d7 10\u207b\u00b3 mol<br>Total atoms = 0.5 \u00d7 10\u207b\u00b3 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 3.011 \u00d7 10\u00b2\u2070<br>Total neutrons = 8 \u00d7 3.011 \u00d7 10\u00b2\u2070 = 2.41 \u00d7 10\u00b2\u00b9<br>Mass of 1 neutron = 1.675 \u00d7 10\u207b\u00b2\u2077 kg<br>Total mass = 2.41 \u00d7 10\u00b2\u00b9 \u00d7 1.675 \u00d7 10\u207b\u00b2\u2077 = 4.04 \u00d7 10\u207b\u2076 kg<\/li>\n\n\n\n<li>Molar mass of NH\u2083 = 17 g<br>Moles in 34 mg = (34 \u00d7 10\u207b\u00b3) \/ 17 = 2 \u00d7 10\u207b\u00b3 mol<br>Protons in 1 NH\u2083 = 7 (N) + 3 (H) = 10<br>Total protons = 2 \u00d7 10\u207b\u00b3 \u00d7 6.022 \u00d7 10\u00b2\u00b3 \u00d7 10 = 1.204 \u00d7 10\u00b2\u00b2<br>Mass of 1 proton = 1.6726 \u00d7 10\u207b\u00b2\u2077 kg<br>Total mass = 1.204 \u00d7 10\u00b2\u00b2 \u00d7 1.6726 \u00d7 10\u207b\u00b2\u2077 = 2.01 \u00d7 10\u207b\u2075 kg<\/li>\n\n\n\n<li>\u2086\u00b9\u00b3C \u2192 Protons = 6, Neutrons = 13 \u2212 6 = 7<br>\u2088\u00b9\u2076O \u2192 Protons = 8, Neutrons = 16 \u2212 8 = 8<br>\u2081\u2082\u00b2\u2074Mg \u2192 Protons = 12, Neutrons = 24 \u2212 12 = 12<br>\u2082\u2086\u2075\u2076Fe \u2192 Protons = 26, Neutrons = 56 \u2212 26 = 30<br>\u2083\u2088\u2078\u2078Sr \u2192 Protons = 38, Neutrons = 88 \u2212 38 = 50<br>Changing temperature and pressure has no effect on the number of protons and neutrons.<\/li>\n\n\n\n<li>(i) Z = 17, A = 35 \u2192 \u2081\u2087\u00b3\u2075Cl<br>(ii) Z = 92, A = 233 \u2192 \u2089\u2082\u00b2\u00b3\u00b3U<br>(iii) Z = 4, A = 9 \u2192 \u2084\u2079Be<\/li>\n\n\n\n<li>Wavelength \u03bb = 580 nm = 580 \u00d7 10\u207b\u2079 m<br>Speed of light c = 3.0 \u00d7 10\u2078 m\/s<br>Frequency \u03bd = c \/ \u03bb = 3.0 \u00d7 10\u2078 \/ 580 \u00d7 10\u207b\u2079 = 5.17 \u00d7 10\u00b9\u2074 Hz<br>Wavenumber \u03bd\u0305 = 1 \/ \u03bb (in cm) = 1 \/ (580 \u00d7 10\u207b\u2077) = 1.72 \u00d7 10\u2074 cm\u207b\u00b9<\/li>\n\n\n\n<li>(i) E = h\u03bd = 6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u00b9\u2075 = 1.9878 \u00d7 10\u207b\u00b9\u2078 J<br>(ii) \u03bb = 0.50 \u00c5 = 0.50 \u00d7 10\u207b\u00b9\u2070 m<br>E = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (0.50 \u00d7 10\u207b\u00b9\u2070) = 3.976 \u00d7 10\u207b\u00b9\u2075 J<\/li>\n\n\n\n<li>T = 2.0 \u00d7 10\u207b\u00b9\u2070 s<br>Frequency \u03bd = 1 \/ T = 5.0 \u00d7 10\u2079 Hz<br>\u03bb = c \/ \u03bd = 3.0 \u00d7 10\u2078 \/ 5.0 \u00d7 10\u2079 = 0.06 m<br>Wavenumber \u03bd\u0305 = 1 \/ \u03bb (in cm) = 1 \/ (6.0 \u00d7 10\u207b\u00b9) = 1.67 cm\u207b\u00b9<\/li>\n\n\n\n<li>Energy of one photon (E) = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (4000 \u00d7 10\u207b\u00b9\u00b2) = 4.97 \u00d7 10\u207b\u00b2\u2070 J<br>Number of photons = 1 \/ 4.97 \u00d7 10\u207b\u00b2\u2070 = 2.01 \u00d7 10\u00b9\u2079<\/li>\n\n\n\n<li>(i) E = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (4 \u00d7 10\u207b\u2077) = 4.97 \u00d7 10\u207b\u00b9\u2079 J<br>Convert to eV: E = 4.97 \u00d7 10\u207b\u00b9\u2079 \/ 1.602 \u00d7 10\u207b\u00b9\u2079 = 3.10 eV<br>(ii) KE = 3.10 \u2212 2.13 = 0.97 eV<br>(iii) KE (in J) = 0.97 \u00d7 1.602 \u00d7 10\u207b\u00b9\u2079 = 1.554 \u00d7 10\u207b\u00b9\u2079 J<br>v = \u221a(2 \u00d7 KE \/ m) = \u221a(2 \u00d7 1.554 \u00d7 10\u207b\u00b9\u2079 \/ 9.109 \u00d7 10\u207b\u00b3\u00b9) = 5.86 \u00d7 10\u2075 m\/s<\/li>\n\n\n\n<li>\u03bb = 242 nm = 242 \u00d7 10\u207b\u2079 m<br>E (per photon) = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (242 \u00d7 10\u207b\u2079) = 8.21 \u00d7 10\u207b\u00b9\u2079 J<br>E (per mole) = 8.21 \u00d7 10\u207b\u00b9\u2079 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 494.5 kJ\/mol<\/li>\n\n\n\n<li>P = 25 W = 25 J\/s<br>\u03bb = 0.57 \u00b5m = 5.7 \u00d7 10\u207b\u2077 m<br>E per photon = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (5.7 \u00d7 10\u207b\u2077) = 3.49 \u00d7 10\u207b\u00b9\u2079 J<br>Number of photons\/sec = 25 \/ 3.49 \u00d7 10\u207b\u00b9\u2079 = 7.16 \u00d7 10\u00b9\u2079<\/li>\n\n\n\n<li>\u03bb = 6800 \u00c5 = 6800 \u00d7 10\u207b\u00b9\u2070 m<br>\u03bd = c \/ \u03bb = 3 \u00d7 10\u2078 \/ 6.8 \u00d7 10\u207b\u2077 = 4.41 \u00d7 10\u00b9\u2074 Hz<br>\u03bd\u2080 = 4.41 \u00d7 10\u00b9\u2074 Hz<br>W\u2080 = h\u03bd\u2080 = 6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 4.41 \u00d7 10\u00b9\u2074 = 2.92 \u00d7 10\u207b\u00b9\u2079 J<\/li>\n\n\n\n<li>For transition n = 4 to n = 2 in hydrogen:<br>1 \/ \u03bb = R (1\/2\u00b2 \u2212 1\/4\u00b2) = 1.097 \u00d7 10\u2077 \u00d7 (1\/4 \u2212 1\/16) = 1.097 \u00d7 10\u2077 \u00d7 3\/16 = 2.056 \u00d7 10\u2076 m\u207b\u00b9<br>\u03bb = 1 \/ 2.056 \u00d7 10\u2076 = 486.1 nm<\/li>\n\n\n\n<li>Energy at n = 5 = \u22122.18 \u00d7 10\u207b\u00b9\u2078 \/ 5\u00b2 = \u22128.72 \u00d7 10\u207b\u00b2\u2070 J<br>Energy required to ionise = 0 \u2212 (\u22128.72 \u00d7 10\u207b\u00b2\u2070) = 8.72 \u00d7 10\u207b\u00b2\u2070 J<br>Ionisation enthalpy from n = 1 = 2.18 \u00d7 10\u207b\u00b9\u2078 J<br>Energy required from n = 5 is much smaller.<\/li>\n\n\n\n<li>Maximum number of emission lines from n = 6 to n = 1 = n(n \u2212 1)\/2 = 6(6 \u2212 1)\/2 = 15<\/li>\n\n\n\n<li>Energy of first orbit = \u22122.18 \u00d7 10\u207b\u00b9\u2078 J<br>Energy of fifth orbit = \u22122.18 \u00d7 10\u207b\u00b9\u2078 \/ 5\u00b2 = \u22128.72 \u00d7 10\u207b\u00b2\u2070\u00a0<\/li>\n\n\n\n<li>The longest wavelength in the Balmer series is for n = 3 to n = 2<br>1 \/ \u03bb = R (1\/2\u00b2 \u2212 1\/3\u00b2) = 1.097 \u00d7 10\u2077 \u00d7 (5\/36) = 1<\/li>\n\n\n\n<li>\u03bb = h \/ (mv) = (6.626 \u00d7 10\u207b\u00b3\u2074) \/ (9.1 \u00d7 10\u207b\u00b3\u00b9 \u00d7 2.05 \u00d7 10\u2077) = 3.55 \u00d7 10\u207b\u00b9\u00b9 m<\/li>\n\n\n\n<li>K.E. = 3.0 \u00d7 10\u207b\u00b2\u2075 J<br>p = \u221a(2mK.E.) = \u221a(2 \u00d7 9.1 \u00d7 10\u207b\u00b3\u00b9 \u00d7 3.0 \u00d7 10\u207b\u00b2\u2075)<br>\u03bb = h \/ p = 6.626 \u00d7 10\u207b\u00b3\u2074 \/ \u221a(5.46 \u00d7 10\u207b\u2075\u2075) = 9.66 \u00d7 10\u207b\u00b9\u2070 m<\/li>\n\n\n\n<li>Isoelectronic species (same electrons):<br>Na\u207a (10), Mg\u00b2\u207a (10), F\u207b (10), O\u00b2\u207b (10), Ne (10),<br>K\u207a (18), Ca\u00b2\u207a (18), S\u00b2\u207b (18), Ar (18)<br>Bohr radius (r\u2099) = 0.529 \u00d7 n\u00b2 = 0.529 \u00d7 25 = 13.23 \u00c5 = 1323 pm<\/li>\n\n\n\n<li>\u00a0(a) H\u207b: 1s\u00b2<br>(b) Na\u207a: 1s\u00b2 2s\u00b2 2p\u2076<br>(c) O\u00b2\u207b: 1s\u00b2 2s\u00b2 2p\u2076<br>(d) F\u207b: 1s\u00b2 2s\u00b2 2p\u2076<br>(ii) (a) Z = 11 (b) Z = 7 (c) Z = 17<br>(iii) (a) Li (b) P (c) Sc<\/li>\n\n\n\n<li>\u00a0g-orbitals: l = 4, so minimum n = 5<\/li>\n\n\n\n<li>\u00a0For 3d orbital: n = 3, l = 2, m\u2097 = \u20132, \u20131, 0, 1, 2<\/li>\n\n\n\n<li>\u00a0(i) 29 protons<br>(ii) Electronic configuration: [Ar] 3d\u00b9\u2070 4s\u00b9<\/li>\n\n\n\n<li>\u00a0(i) l = 0, 1, 2; m\u2097 = \u2212l to +l<br>(ii) 3d \u2192 l = 2, m\u2097 = \u22122, \u22121, 0, 1, 2<br>(iii) 1p , 2s , 2p , 3f\u00a0<\/li>\n\n\n\n<li>\u00a0(a) n=1, l=0 \u2192 1s<br>(b) n=3, l=1 \u2192 3p<br>(c) n=4, l=2 \u2192 4d<br>(d) n=4, l=3 \u2192 4f<\/li>\n\n\n\n<li>\u00a0(a) n=4, m\u209b = \u2212\u00bd \u2192 8 electrons<br>(b) n=3, l=0 \u2192 2 electrons<\/li>\n\n\n\n<li>\u00a02\u03c0r = n\u03bb \u2192 Bohr\u2019s orbit is an integral multiple of de Broglie wavelength.<\/li>\n\n\n\n<li>\u00a0Transition n=4 to n=2 in He\u207a is the same as n=8 to n=4 in H.<\/li>\n\n\n\n<li>\u00a0Energy required = Z\u00b2 \u00d7 2.18 \u00d7 10\u207b\u00b9\u2078 = 4 \u00d7 2.18 \u00d7 10\u207b\u00b9\u2078 = 8.72 \u00d7 10\u207b\u00b9\u2078 J<\/li>\n\n\n\n<li>\u00a0No. of atoms = 0.2 m \/ 1.5 \u00d7 10\u207b\u00b9\u2070 m = 1.33 \u00d7 10\u2079<\/li>\n\n\n\n<li>\u00a0Radius = 2.4 \/ (2 \u00d7 10\u2078) = 1.2 \u00d7 10\u207b\u2078 m = 120 pm<\/li>\n\n\n\n<li>\u00a0Radius = 1.3 \u00c5 = 130 pm<br>Atoms = 1.6 \u00d7 10\u207b\u00b2 m \/ 2.6 \u00d7 10\u207b\u00b9\u2070 m = 6.15 \u00d7 10\u2077<\/li>\n\n\n\n<li>\u00a0Electrons = Q \/ e = 2.5 \u00d7 10\u207b\u00b9\u2076 \/ 1.6 \u00d7 10\u207b\u00b9\u2079 = 1562.5 \u2248 1.56 \u00d7 10\u00b3<\/li>\n\n\n\n<li>\u00a0Electrons = 1.282 \u00d7 10\u207b\u00b9\u2078 \/ 1.6 \u00d7 10\u207b\u00b9\u2079 = 8.01 \u2248 8 electrons<\/li>\n\n\n\n<li>\u00a0Light atoms deflect \u03b1-particles more, with fewer large-angle deflections than with heavy atoms.<\/li>\n\n\n\n<li>\u00a0Only \u2077\u2079\u2083\u2085Br is correct. Just \u2077\u2079Br or \u00b3\u2075Br are incomplete (lack atomic\/mass number).<\/li>\n\n\n\n<li>\u00a0Let protons = x \u2192 neutrons = x + 0.317x = 1.317x<br>x + 1.317x = 81 \u2192 x = 34.9 \u2192 Z = 35, so symbol = \u2078\u00b9\u2083\u2085Br<\/li>\n\n\n\n<li>\u00a0Let electrons = x \u2192 neutrons = x + 0.111x = 1.111x<br>x + 1.111x = 36, x = 17.05 \u2192 electrons = 17, protons = 18<br>Symbol = \u00b3\u2077\u2081\u2088Ar\u207b<\/li>\n\n\n\n<li>\u00a0x + 1.304x = 56 \u2192 x = 24.2 \u2192 electrons = 24, protons = 27<br>Symbol = \u2075\u2076\u2082\u2087Co\u00b3\u207a<\/li>\n\n\n\n<li>\u00a0Frequency order (increasing):<br>FM radio < microwave oven < amber light < X-rays < cosmic rays<\/li>\n\n\n\n<li>\u00a0E per photon = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (337.1 \u00d7 10\u207b\u2079) = 5.9 \u00d7 10\u207b\u00b9\u2079 J<br>Total energy = 5.6 \u00d7 10\u00b2\u2074 \u00d7 5.9 \u00d7 10\u207b\u00b9\u2079 = 3.3 \u00d7 10\u2076 J<br>Power = E \/ t = 3.3 \u00d7 10\u2076 W<\/li>\n\n\n\n<li>\u00a0(a) \u03bd = c \/ \u03bb = 3 \u00d7 10\u2078 \/ 616 \u00d7 10\u207b\u2079 = 4.87 \u00d7 10\u00b9\u2074 Hz<br>(b) Distance = c \u00d7 t = 3 \u00d7 10\u2078 \u00d7 30 = 9 \u00d7 10\u2079 m<br>(c) E = hc \/ \u03bb = 3.22 \u00d7 10\u207b\u00b9\u2079 J<br>(d) Quanta = 2 \/ 3.22 \u00d7 10\u207b\u00b9\u2079 = 6.2 \u00d7 10\u00b9\u2078<\/li>\n\n\n\n<li>\u00a0E = hc \/ \u03bb = (6.626 \u00d7 10\u207b\u00b3\u2074 \u00d7 3 \u00d7 10\u2078) \/ (600 \u00d7 10\u207b\u2079) = 3.313 \u00d7 10\u207b\u00b9\u2079<br>Photons = 3.15 \u00d7 10\u207b\u00b9\u2078 \/ 3.313 \u00d7 10\u207b\u00b9\u2079 = 9.51 photons<\/li>\n\n\n\n<li>\u00a0E per photon = hc \/ \u03bb = 5.91 \u00d7 10\u207b\u00b9\u2079<br>Total energy = 2.5 \u00d7 10\u00b9\u2075 \u00d7 5.91 \u00d7 10\u207b\u00b9\u2079 = 1.48 \u00d7 10\u207b\u00b3 J<\/li>\n\n\n\n<li>\u00a0\u03bd\u2081 = 3 \u00d7 10\u2078 \/ 589 \u00d7 10\u207b\u2079 = 5.09 \u00d7 10\u00b9\u2074<br>\u03bd\u2082 = 3 \u00d7 10\u2078 \/ 589.6 \u00d7 10\u207b\u2079 = 5.087 \u00d7 10\u00b9\u2074<br>\u0394E = h(\u03bd\u2082 \u2212 \u03bd\u2081) = 1.06 \u00d7 10\u207b\u00b2\u2070 J<\/li>\n\n\n\n<li>\u00a0(a) \u03bb = hc \/ E = 6.56 \u00d7 10\u207b\u2077 m<br>(b) \u03bd = c \/ \u03bb = 4.57 \u00d7 10\u00b9\u2074 Hz<br>E = hc \/ \u03bb = 3.98 \u00d7 10\u207b\u00b9\u2079<br>KE = 3.98 \u2212 1.9 = 2.08 eV<br>v = \u221a(2 \u00d7 KE \u00d7 e \/ m) = 8.52 \u00d7 10\u2075 m\/s<\/li>\n\n\n\n<li>\u00a0v\u00b2 = 2KE \/ m = gives threshold KE = 0 when \u03bb = 688 nm<br>Use E = hc \/ \u03bb, find h = (KE \u00d7 \u03bb) \/ c \u2192 h = 6.626 \u00d7 10\u207b\u00b3\u2074 J\u00b7s<\/li>\n\n\n\n<li>\u00a0KE = eV = 0.35 eV = 5.6 \u00d7 10\u207b\u00b2\u2070<br>E = hc \/ \u03bb = 7.74 \u00d7 10\u207b\u00b9\u2079<br>Work function = 7.74 \u00d7 10\u207b\u00b9\u2079 \u2212 5.6 \u00d7 10\u207b\u00b2\u2070 = 7.18 \u00d7 10\u207b\u00b9\u2079 J<\/li>\n\n\n\n<li>\u00a0E = hc \/ \u03bb = 1.326 \u00d7 10\u207b\u00b9\u2075<br>KE = \u00bdmv\u00b2 = 1.025 \u00d7 10\u207b\u00b9\u2079<br>Binding energy = E \u2212 KE = 1.223 \u00d7 10\u207b\u00b9\u2075 J<\/li>\n\n\n\n<li>\u00a01285 nm = 1.285 \u00d7 10\u207b\u2076 m<br>\u03bd = c \/ \u03bb = 2.33 \u00d7 10\u00b9\u2074<br>Solve: 3.29 \u00d7 10\u00b9\u2075 (1\/9 \u2212 1\/n\u00b2) = 2.33 \u00d7 10\u00b9\u2074 \u2192 n = 6<br>Region: Infrared<\/li>\n\n\n\n<li>\u00a0R \u221d n\u00b2<br>r\u2081 = 1.3225 nm, r\u2082 = 211.6 pm<br>Find n\u2081 and n\u2082, use E = hc \/ \u03bb<br>Wavelength = 9.75 \u00d7 10\u207b\u2078 m<br>Series = Lyman<br>Region = UV<\/li>\n\n\n\n<li>\u00a0\u03bb = h \/ mv = 6.626 \u00d7 10\u207b\u00b3\u2074 \/ (9.1 \u00d7 10\u207b\u00b3\u00b9 \u00d7 1.6 \u00d7 10\u2076) = 4.54 \u00d7 10\u207b\u00b9\u2070 m<\/li>\n\n\n\n<li>\u00a0\u03bb = 800 pm = h \/ mv \u2192 v = h \/ (m\u03bb) = 9.06 \u00d7 10\u00b2 m\/s<\/li>\n\n\n\n<li>\u00a0\u03bb = h \/ mv = 6.626 \u00d7 10\u207b\u00b3\u2074 \/ (9.1 \u00d7 10\u207b\u00b3\u00b9 \u00d7 2.19 \u00d7 10\u2076) = 3.31 \u00d7 10\u207b\u00b9\u2070 m<\/li>\n\n\n\n<li>\u00a0m (proton) = 1.67 \u00d7 10\u207b\u00b2\u2077 kg<br>\u03bb = h \/ mv = 9.68 \u00d7 10\u207b\u00b9\u2070 m<br>For hockey ball: \u03bb = 1.51 \u00d7 10\u207b\u00b3\u00b3 m<\/li>\n\n\n\n<li>\u00a0\u0394x = 0.002 nm = 2 \u00d7 10\u207b\u00b9\u00b2 m<br>\u0394p \u2265 h \/ (4\u03c0\u0394x) = 2.63 \u00d7 10\u207b\u00b2\u00b3 kg\u00b7m\/s<br>Given p = h \/ (4\u03c0m \u00d7 0.05 nm) = 1.16 \u00d7 10\u207b\u00b2\u00b3, consistent<\/li>\n\n\n\n<li>\u00a0Energy order (increasing):<br>(5) < (2), (4) < (3), (6) = degenerate<\/li>\n\n\n\n<li>\u00a0Least Z_eff felt by 4p electron (furthest from nucleus)<\/li>\n\n\n\n<li>\u00a0(i) 2s > 3s \u2192 2s feels more Z_eff<br>(ii) 4d > 4f \u2192 4d feels more<br>(iii) 3p > 3d \u2192 3p feels more<\/li>\n\n\n\n<li>\u00a0Si 3p electron feels more Z_eff due to the higher atomic number<\/li>\n\n\n\n<li>(a) P = 3, Si = 2, Cr = 6, Fe = 4, Kr = 0<br>(b) Subshells for n=4: 4s, 4p, 4d, 4f \u2192 4 subshells<br>Each orbital holds 2 electrons, half with m\u209b = \u2212\u00bd: 16 electrons<\/li>\n<\/ol>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-geography-fundamentals-of-physical-geography-chapter-11-world-climate-and-climate-change\/\"><strong>NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)<\/strong><\/a><\/strong><\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-download-ncert-solutions-class-11-chemistry-part-i-chapter-2-structure-of-atom\">Download<strong> <\/strong>NCERT Solutions Class 11 Chemistry (Part-I) Chapter 2: Structure of Atom<\/h2>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#cd9ef9\"><tbody><tr><td><strong><a href=\"https:\/\/drive.google.com\/file\/d\/1lKmamwqxY5Lc6uuOOTPrnXd_TnRDYA9Z\/view?usp=drive_link\">Download PDF of NCERT Solutions Class 11 Chemistry (Part-1) Chapter 2: Structure of Atom<\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Download the Solutions of Other Chapters of Class 11 Psychology<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f4e0a9\"><tbody><tr><td><strong>Chapter <\/strong>1<\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><td><strong>Chapter 6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Related Reads<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-pale-ocean-gradient-background has-background has-fixed-layout\"><tbody><tr><td><strong><a 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href=\"https:\/\/leverageedu.com\/discover\/category\/school-education\/ncert-study-material\/\"><strong>NCERT Study Material<\/strong><\/a> today!\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"In this unit, we have earlier provided you with a summary of notes on the structure of atoms&hellip;\n","protected":false},"author":133,"featured_media":866210,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-866208","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.5 (Yoast SEO v27.5) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 11 Chemistry Chapter-2: Structure of Atom (Free 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