Question

provide 1 missing symbol and 4 numbers 114m b c in → in + a d e o a = 1 b = 2 c = 3 d = 4 e = 5 a. γ (gamma) b. 0 c. 232 d. 90 e. he f. th g. 4 h. 88 i. 188 j. 114 k. re l. -1 m. e n. bi o. pb p. 82 q. 208 r. 248 s. sn t. 7

Explanation:

Step1: Identify nuclear - decay type

This seems to be a nuclear - decay equation. Since the parent and daughter nuclides are the same (In), and only a particle is emitted, it is likely a gamma - decay as gamma decay does not change the mass number and atomic number of the nuclide. So A is $\gamma$ (gamma).

Step2: Analyze mass number and atomic number conservation

In nuclear reactions, mass number and atomic number are conserved. For indium (In), the atomic number is 49. Since it is gamma decay, the mass number and atomic number of the products remain the same as the reactant. There is no change in mass number and atomic number during gamma decay. So B = 0 (change in atomic number), C = 0 (change in mass number), D = 0 (change in atomic number on the left - hand side for the particle being emitted), E = 0 (change in mass number on the left - hand side for the particle being emitted).

Answer:

A. $\gamma$ (gamma)
B. 0
C. 0
D. 0
E. 0