236Write a nuclear equation for the alpha decay ofPu.94236A.94232Am95Pu >e +-1232236В.944Pu >Не +292236С.94236Np93Pu >e ++1236235D.Pu →n +Pu9494O A. Reaction DB. Reaction BO C. Reaction AO D. Reaction C

In alpha decay radioactive element goes into disintegration and emitts alpha particles.The correct answer is B that is 236 94Pu --------> 4 2He + 23292U In alpha decay unstable nucleus goes into disintegration to emiit alpha particles that is helium nucleus. In alpha decay atomic number is reduced by two and atomic mass is reduced by four therefore the correct answer is B.

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Chemistry

236 Write a nuclear equation for the alpha decay of Pu. 94 236 pu A. 94 232 e + Am -1 95 232 236 В. 94 4 Pu > Не + 2 92 236 С. 94 236 Np 93 Pu > e + +1 236 235 D. Pu → n + Pu 94 94 O A. Reaction D B. Reaction B O C. Reaction A O D. Reaction C

236 Write a nuclear equation for the alpha decay of Pu. 94 236 pu A. 94 232 e + Am -1 95 232 236 В. 94 4 Pu > Не + 2 92 236 С. 94 236 Np 93 Pu > e + +1 236 235 D. Pu → n + Pu 94 94 O A. Reaction D B. Reaction B O C. Reaction A O D. Reaction C

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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