Question

provide 3 missing symbols and 1 number
208 0 208
a + b_orbital → c
83 -1 d
a = 1
b = 2
c = 3
d = 4
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 u. f
v. n w. 19 x. 228 y. ra z. 2

Explanation:

Step1: Identify the element with atomic number 83

The element with atomic number 83 is Bismuth, symbol Bi. So A = N. Bi

Step2: Identify the symbol for an electron in an orbital

The symbol for an electron is e. So B = M. e

Step3: Calculate the mass - number and atomic - number conservation

In a nuclear reaction, the sum of mass - numbers and atomic - numbers on both sides of the equation must be equal. The mass - number of A (Bi) is 208, and the mass - number of the electron is 0. So the mass - number of C is 208 + 0=208. The atomic - number of A (Bi) is 83, and the atomic - number of the electron is - 1. So the atomic - number of C is 83+( - 1)=82. The element with atomic number 82 is Lead, symbol Pb. So C = O. Pb

Step4: Determine the atomic - number of C

The atomic - number of C (Pb) is 82. So D = P. 82

Answer:

1. N. Bi
2. M. e
3. O. Pb
4. P. 82