Question

provide 2 missing symbols and 3 numbers
a 110 134 d
cm → ru + b + c
96 44 50 e
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 u. f

Explanation:

Step1: Identify element symbols and atomic numbers

This is a nuclear - decay equation. Cm (Curium) has atomic number 96. Ru (Ruthenium) has atomic number 44.

Step2: Balance atomic numbers

For the atomic - number balance in the nuclear reaction \(_{96}^{A}Cm
ightarrow_{44}^{110}Ru + _{Z_1}^{134}B+_{Z_2}^{D}C\), we know that \(96 = 44+Z_1 + Z_2\).

Step3: Balance mass numbers

For the mass - number balance, \(A=110 + 134+D\).

Step4: Analyze options

By looking at the options and considering nuclear - decay rules, in a nuclear reaction, if we assume \(B\) is \(_{50}^{134}Sn\) (since \(Z_1 = 50\) to balance atomic numbers as \(96-(44 + 50)=2\)) and \(C\) is \(_{2}^{4}He\) (alpha - particle, which is common in nuclear decays and \(Z_2 = 2\)). Then \(A=110 + 134+4=248\) and \(D = 4\). Also, for the energy - related part, a gamma ray \(\gamma\) (option A) can be emitted in nuclear reactions. And if we consider conservation of other quantum numbers, we can fill the blanks.

Answer:

A. 248
B. Sn
C. He
D. 4
E. \(\gamma\) (gamma)