Important Questions -Atoms & Nuclei

Important questions – Atoms & Nuclei 1 When is Ha line of the Balmer series in the emission spectrum of hydrogen atom obtained? Ha line of the Balmer series in the emission spectrum of hydrogen atom is obtained in the visible region. When electron jumps from n = 2 to ni = 3, then Ha line will be obtained. 2 What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the fourth excited state? 3 The ground state energy of hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 1.51 eV to – 3.4 eV, calculate the wavelength of the spectral line emitted and the series of hydrogen spectrum to which it belongs 4 The short wavelength limits of Lyman, Paschen and Balmer series in the hydrogen spectrum are denoted by λL, λP and λB respectively. Arrange these wavelengths in increasing order. 5 The nucleus of an atom of Y235 initially at rest decays by emitting an α- particle. The binding energy per nucleon 92 of parent and daughter nuclei are 7.8MeV and 7.835MeV respectively and that of α- particles is 7.07MeV/nucleon. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of α- particle to be 6.68×10−27kg. Page 1 of 14

6 Find the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmer and Paschen series of the hydrogen spectrum. 7 A star converts all its hydrogen to helium, achieving 100% helium composition. It then converts the helium to carbon via the reaction 2He4+2He4+2He4→6C12+7.27MeV The mass of the star is 5.0×1032 kg and it generates energy at the rate of 5×1030kg watt. How long will it take to convert all the helium to carbon at this rate? Page 2 of 14

8 (i) If electron in the atom is replaced by a particle(muon) having the same charge but mass about 200 times as that of the electron to form a muonic atom, how would : (i) the radius and (ii) the ground state energy of this be affected ? (ii) Calculate the wavelength of the first spectral line in the corresponding Lyman series of this atom. 9 n=2 Page 3 of 14

10 Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer. In absence of magnetic field, the energy is determined by the principle quantum number n, while l is the orbital quantum number. If an electron is in nth state then the magnitude of the angular momentum is where, l = 0, 1, 2, .........., (n – 1) Since l = 0, 1, 2, ....., (n – 1), different values of l are compatible with the same value of n. For example, when n = 3, the possible values of l are 0, 1, 2, and when n = 4, the possible values of l are 0, 1, 2, 3. Thus, the electron in one of the atoms could have n = 3, l = 2, while the electron in the other atom could have n = 4, l = 2. Therefore, according to quantum mechanics, it is possible for the electrons to have different energies but have the same orbital angular momentum. 11. (i) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain. (ii) Plot a graph showing variation of potential energy of a pair of nucleons as a function of their separation. (i) Protons and neutrons have to come very near to bind together and form a nucleus. This distance is of the order of 10–14 m. In order to achieve this distance a lot of energy is required. The required energy is provided by the nucleon at the expense of some portion of their masses. This is the reason that the mass of a nucleus in its ground state is always less than the total mass of its constituents –neutrons and protons. 12 Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level. (All India 2010) Answer: 13 Define the distance of closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is V. What will be the distance of closest approach for an a-particle of double the kinetic energy? (Delhi 2016) Answer: The distance of closest approach is defined as “the distance of charged particle from the centre of the nucleus, at which the whole of the initial kinetic energy of the (far off) charged particle gets converted into the electric potential energy of the system”. Distance of closest approach (rc) is given by 14 The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of electron in this state? (All India) Answer: Kinetic energy, Ke = + T.E. = 13.6 eV Potential energy, Pe = 2 T.E. = 2 (-13.6) = – 27.2 eV Page 4 of 14

15 (i) In hydrogen atom, an electron undergoes transition from 2nd excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong. (ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases. (Comptt. All India 2012) Answer: 16 Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie? (All India 2012) Answer: In Balmer series, an electron jumps from higher orbits to the second stationary orbit (nf = 2). Thus for this series : 17 The figure shows energy level diagram of hydrogen atom Page 5 of 14

(a) Find out the transition which results in the emission of a photon of wavelength 496 nm. (b) Which transition corresponds to the emission of radiation of maximum wavelength? Justify your answer. (Comptt. All India 2012) Answer: (a) Transition emitting wavelength λ = 496 nm The given wavelength lies in visible region (Balmer series) when, which means that the maximum wavelength emmission will be there when the energy level difference is minimum. From the given energy level diagram, it corresponds to : 18 Find the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmer and Paschen series of the hydrogen spectrum. (Comptt. All India 2016) Answer: 19 Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii? (All India 2009) Answer: Page 6 of 14

20 A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted. (All India 2016) Answer: 21 The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A Calculate the short wavelength limit for Balmer series of the hydrogen spectrum. (All India 2016) Answer: Page 7 of 14

22 Calculate the longest wavelength of the photons emitted in the Balmer series of hydrogen spectrum. Which part of the e.m. spectrum, does it belong to? [Given Rydberg constant, R = 1.1 × 107 m-1]. (Comptt. Delhi 2016) Answer: Balmer series is produced when an electron jumps from higher orbits to second stationary orbit (nf = 2). 23 The energy level diagram of an element is given. Identify, by doing necessary calculations, which transition corresponds to the emission of a spectral line of wavelength 102.7 nm. (Delhi 2008) Answer: 24 The ground state energy of hydrogen atom is -13.6 eV. (i) What is the potential energy of an electron in the 3rd excited state? (ii) If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted. (All India 2008) Answer: Page 8 of 14

25 State the reason, why heavy water is generally used as a moderator in a nuclear reactor. (Delhi 2008) Answer: Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction : ere d is deutron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross- section for neutron absorption. 26 A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV. (Delhi 2016) Answer: ∴ Gain in binding energy for nucleon = 8.5 – 7.6 = 0.9 MeV Hence total gain in binding energy per nucleus fission = 240 × 0.9 = 216 MeV 27 Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. (All India 2016) Answer: Two important conclusions : (i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large-size nucleus. (ii) Graph explains that force is attractive for distances larger than 0.8 fin and repulsive for distances less than 0.8 fm. Page 9 of 14

28 (i) State Bohr’s quantization condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits? (ii) Find the relation between the three wave-lengths λ1, λ2 and λ3 from the energy level diagram shown below: (Delhi 2016) Answer: Page 10 of 14

29 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Page 11 of 14

31 (A) What is the potential energy of an electron in the 3rd excited state? (B) If the electron jump to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted. 32 Page 12 of 14

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33 The spectrum of a star in the visible and the ultraviolet region was observed and the wavelength of some of the lines that could be identified was found to be: 824A0,970A0,1120A∘,2504A0,5173A0,6100A0 Which of these lines cannot belong to the hydrogen atom spectrum? (Given Rydberg constant R=1.03×107m−1 and R1=970A0. Support your answer with suitable calculations Page 14 of 14