Help please Using your knowledge of alpha beta and Gamma decay and your periodic table practice using your ideas by completing the following equations

20. ²⁴¹₉₅Am —> ²³⁷₉₃Np + ⁴₂He

21. ⁴²₁₉K —> ⁰₋₁β + ⁴²₂₀Ca

Explanation:

20. ²⁴¹₉₅Am —> __ + ⁴₂He

Let ⁿₘX be the unknown.

Thus, the equation becomes:

²⁴¹₉₅Am —> ⁿₘX + ⁴₂He

Next, we shall determine n, m and X. This can be obtained as follow:

241 = n + 4

Collect like terms

n = 241 – 4

n = 237

95 = m + 2

Collect like terms

m = 95 – 2

m = 93

Thus,

ⁿₘX => ²³⁷₉₃X => ²³⁷₉₃Np

Therefore, the complete equation is:

²⁴¹₉₅Am —> ²³⁷₉₃Np + ⁴₂He

21. __ —> ⁰₋₁β + ⁴²₂₀Ca

Let ʸₓZ be the unknown.

Thus, the equation becomes:

ʸₓZ —> ⁰₋₁β + ⁴²₂₀Ca

Next, we shall determine y, x and Z. This can be obtained as follow:

y = 0 + 42

y = 42

x = –1 + 20

x = 19

Thus,

ʸₓZ => ⁴²₁₉Z => ⁴²₁₉K

Therefore, the complete equation is:

⁴²₁₉K —> ⁰₋₁β + ⁴²₂₀Ca

the correct answer would be d. hope this !