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Important Problems of Velocity and Energy of electron (In Hindi)
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In this lesson we will learn about some important problems based on the concept of velocity of electron and energy of electron in nth orbit for IIT-JEE students

Niranjay Kumar Dwivedi
(N.K.D.Sir) is a Complete Chemistry faculty for IIT~JEE, and NEET examination. with more then 8 year of Teaching experience

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Mahesh Mishra
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1. Problem OF Velocity & energy of electrons in nth orbit (Bohrs atomic model) By-N.K.D. Sir Please -Rate review and share this lesson

2. The ionization energy of H-atom is 13.6 eV. What will be ionization energy of Het and Li2+ ions? Solution IE1 for He+ 1E1 for H x Z2 13.6 x 4 54.4 eV Also, IEl for Li?+ = 1E1 for H Z2 13.6 x 9122.4 eV The ionization energy of He+ is 19.6 10-18 J per atom. Calculate the energy of the second stationary state in Li2+ ion Solution The energy of the first orbit in one electron species is related to that in hydrogen atom by an expression For He+ ion, z-2. Hence, E 19.6x1018 4 Now for Li2+ ion, where Z -3, we will have E = (9)E,, = 91-19.6x10-18 44.1x10 18 J 4 Now for the second orbit, the energy is -44 1 10-18J 2 E =-11.025 x10-18.1

3. The binding energy of an electron in the ground state of the He atom is equal to 24.6 eV. The energy required to remove both the electrons from the atom will be (A) 59 eV (C) 79 eV (B) 81 eV (D) None of these Solution: Ionization energy of He h2. 13.6 22 He = 54.4 eV Energy required to remove both the electrons = binding energy + ionization energy = 24.6 + 54.4 =79 eV -- 13.6

4. How many chlorine atoms can you ionize in the process Cl-> cr+ e, by the energy liberated from the following process. Cl +eci for 6 x 1023 atoms Given electron affinity of Cl = 3.61 eV, and i P of Cl = 17.422 eV (A) 1.24 x1028 atoms (C) 2.02 x 1015 atoms (B) 9.82 x1020 atoms (D) none of these Solution: Energy released in conversion of 6 x 1023 atoms of Cl- ions 6 1023 electron affinity 6 1023 3.61 = 2.166 1024 eV Let x Cl atoms are converted to Cl ion Energy absorbed = x ionization energy x x 17.422 2.166 x 1024 x = 1.243 1023 atoms

5. The speed of the electron in the 1st orbit of the hydrogen atom in the ground state is [C is the velocity of light] (A)-e (B) 1.37 1370 13.7 13.7 h 2nmr 2.189 x 108 cm sec-1 Solution: V- C 3 1010 cm, 137 The ionization energy of the ground state hydrogen atom is 2.18x10-18J. The energy of an electron in its second orbit would be (A)-1.09 x10-18J (C) -4.36 x10-18] (B) -2.18 x10-18 (D) -5.45 x10-19] Solution: Energy of electron in first Bohr's orbit of H-atom E 2.18 10 E2- 2.18x101 -5.45 x10-19J ionization energy of H 2.18 x10-18) 2

6. The velocity of electron in the ground state of H atom is 2.184 x 108 cm/sec. The velocity of electron in the first excited state of Li2*ion in cm/sec would be 3.276 x 108(B) 4.91 x 108 (D) 2.184 x 108 1.638 x 108 Solution: Vn(VIH velocity in the ground state of hydrogen atom) First excited state is n 2, Z - 3 for Li+2 ion 2.184 x 10 x3 -3.276 x 10 2 The ratio of the electron in second excited state of He+ ion to the electron in the first excited state of Be3+ is 1:3 (B) 9:16 1:9 (D) 16:9 2 En =-136 Z 2nd excited state of Het n 3, Z-2 for He* E3--13.6x (2)"--13.6x4 1st excited state of Be3+ is n 2 and Z 4 for Be3+ Solution: 13.6x4213.6 x16 E 4x41 E2 16x9 9 4