Ch09 Atomic Model and Nuclear Physics

PHYSICS 2ND PAPER CHAPTER 9 Atomic Model and Nuclear Physics Question 1 In a hydrogen atom, an electron of mass 𝑚𝑒= 9.11 × 10−31 kg and charge 𝑒= 1.6 × 10−19 C revolves around a central proton (𝑍= 1) in stable circular orbits. The Rydberg constant is 𝑅𝐻= 1.09677 × 107 m−1 and the speed of light in vacuum is 𝑐= 3.0 × 108 m s−1. (a) What is Bohr radius? State its standard numerical value. [R. B. 2023] (b) Explain why the orbital speed of an electron in the 1st orbit of a hydrogen atom is greater than that in the 2nd orbit. [S. B. 2023, C. B. 2022] (c) When an electron in a hydrogen atom jumps from the 2nd energy level to the 1st energy level, calculate the wavelength and frequency of the emitted spectral line. [BUET 2014–2015] (d) An electron in a hydrogen atom makes a transition from energy level 𝑛1 = 4 to some other level 𝑛2. The orbital time period of the electron in the initial state is eight times that in the final state. Find the value of 𝑛2 and determine the orbital radius of this final state. [D. U. 2023–2024] Solution (a) The radius of the innermost stationary circular orbit (𝑛= 1) of the electron in a ground-state hydrogen atom is called the Bohr radius (𝑟1 or 𝑎0). Its theo

(c) Here, Initial energy level 𝑛2 = 2 Final energy level 𝑛1 = 1 Rydberg constant 𝑅𝐻= 1.09677 × 107 m−1 Speed of light 𝑐= 3.0 × 108 m s−1 According to Rydberg’s formula for the hydrogen emission spectrum: 1 𝜆= 𝑅𝐻 1 𝑛2 1 −1 𝑛2 2 ! = 𝑅𝐻  1 12 −1 22  = 𝑅𝐻  1 −1 4  = 3 4 𝑅𝐻 𝜆= 4 3𝑅𝐻 = 4 3 × 1.09677 × 107 m−1 = 1.21569 × 10−7 m = 1215.69 Å The frequency of the emitted photon is: 𝜈= 𝑐 𝜆= 3.0 × 108 m s−1 1.21569 × 10−7 m = 2.4677 × 1015 Hz 𝜆= 1215.69 Å, 𝜈= 2.468 × 1015 Hz (d) The orbital radius and orbital speed for the 𝑛-th orbit are given by: 𝑟𝑛∝𝑛2 and 𝑣𝑛∝1 𝑛 The orbital time period 𝑇𝑛is the circumference divided by orbital speed: 𝑇𝑛= 2𝜋𝑟𝑛 𝑣𝑛 ∝𝑛2 1/𝑛= 𝑛3 Therefore, the ratio of orbital time periods between states 𝑛1 and 𝑛2 is: 𝑇1 𝑇2 = 𝑛1 𝑛2 3 Here, 𝑛1 = 4 and 𝑇1 = 8𝑇2:  4 𝑛2 3 = 8 = 23 =⇒ 4 𝑛2 = 2 =⇒𝑛2 = 2 The radius of the final orbit (𝑛2 = 2) is: 𝑟2 = 𝑛2 2𝑟1 = 22 × 0.529 Å = 4 × 0.529 Å = 2.116 Å = 2.12 × 10−10 m 𝑛2 = 2, 𝑟2 = 2.12 Å ahbab.dev 2

Question 2 Sara and Nuba were conducting research using two distinct radioactive elements. They started counting the number of intact active atoms simultaneously. The variation of their respective intact number of atoms (𝑁) with time 𝑡(in minutes) is shown in the graphs below: 𝑁Element of Sara 1 2 3 4 5 6 𝑡(minute) 84.09 × 106 50 × 106 14.8 × 106 𝑁Element of Nuba 0 1 2 3 4 5 6 𝑡(minute) 100 × 106 65 × 106 48.77 × 106 (a) What is decay constant? [S. B. 2015, All Boards 2018, C. B. 2024] (b) Explain why the termination time (complete disintegration time) of a radioactive substance is theoretically infinite. [R. B. 2016] (c) Determine the decay constant and half-life of the radioactive element of Nuba using data from the graph. [D. B. 2017] (d) Verify in the light of the stem whose element will decay earlier. [D. B. 2017] Solution (a) The decay constant (𝜆) is defined as the fractional disintegration rate per unit time, representing the probability of decay per nucleus per unit time: 𝜆= −1 𝑁 d𝑁 d𝑡 (b) According to the radioactive decay law: 𝑁(𝑡) = 𝑁0𝑒−𝜆𝑡=⇒𝑡= 1 𝜆ln  𝑁0 𝑁  For a radioactive sample to completely disintegrate, the number of undecayed active atoms remaining must be exact