A radioactive isotope has a decay constant of 10^-8 s^-1. Calculate its half-life.

Here are the solutions to the questions: Question 8: A radioactive isotope has a decay constant of 10^-8 s^-1. Calculate its half-life. Step 1: Identify the given decay constant (). = 10^-8 s^-1 Step 2: Use the formula for half-life (T_1/2). T_1/2 = ((2))/() Step 3: Substitute the value of into the formula. T_1/2 = (0.693)/(10^-8) s^-1 Step 4: Calculate the half-life. T_1/2 = 0.693 × 10^8 s T_1/2 = 6.93 × 10^7 s Comparing this to the options: A. 5.93 × 10^6 s B. 6.93 × 10^7 s C. 6.93 × 10^9 s D. 5.93 × 10^7 s The calculated half-life matches option B. The final answer is B --- Question 9: The transverse wave with the longest wavelength is: A. x-rays B. infrared ray C. radio wave D. ultra-violent rays (ultraviolet rays) The electromagnetic spectrum arranges waves by wavelength (or frequency/energy). In order of increasing wavelength, the spectrum is: Gamma rays < X-rays < Ultraviolet rays < Visible light < Infrared rays < Microwaves < Radio waves. From this order, radio waves have the longest wavelength among the given options. The final answer is C --- Question 10: When ^210_82Pb decays to ^206_80Pb, it emits: Step 1: Write the nuclear decay equation. Let x be the number of alpha particles (^4_2He) and y be the number of beta particles (^0_-1e). ^210_82Pb ^206_80Pb + x · ^4_2He + y · ^0_-1e Step 2: Conserve the mass number (superscript). 210 = 206 + 4x + 0y 210 - 206 = 4x 4 = 4x x = 1 This means one alpha particle is emitted. Step 3: Conserve the atomic number (subscript). 82 = 80 + 2x + (-1)y Substitute x=1: 82 = 80 + 2(1) - y 82 = 82 - y y = 0 This means no beta particles are emitted. Therefore, the decay process emits one alpha particle. Comparing this to the options: A. two alpha and two beta particles. B. a beta particle C. one alpha and (assuming this means "one alpha particle") Based on the calculation, one alpha particle is emitted. The final answer is C