Question

Write equations for the following reactions. Beta decay of Cf 256, Es 256, and Md260

Answer 1

1) for Cf-256

Let the product be X

Balance mass number on both sides

mass of Cf = mass of B + mass of X

1*256 = 1*0 + M

M = 256

Balance atomic number on both sides

atomic of Cf = atomic of B + atomic of X

1*98 = 1*-1 + A

A = 99

Atomic number 99 is for element Es

So, the product is

256-Es

Answer:

₉₈²⁵⁶Cf —> ₉₉²⁵⁶Es + _-1⁰B

2) for Es 256

Let the product be X

Balance mass number on both sides

mass of Es = mass of B + mass of X

1*256 = 1*0 + M

M = 256

Balance atomic number on both sides

atomic of Es = atomic of B + atomic of X

1*99 = 1*-1 + A

A = 100

Atomic number 100 is for element Fm

So, the product is

256-Fm

Answer:

₉₉²⁵⁶Es —> ₁₀₀²⁵⁶Fm + _-1⁰B

3)for Md 260

Let the product be X

Balance mass number on both sides

mass of Md = mass of B + mass of X

1*260 = 1*0 + M

M = 260

Balance atomic number on both sides

atomic of Md = atomic of B + atomic of X

1*101 = 1*-1 + A

A = 102

Atomic number 102 is for element No

So, the product is

260-No

Answer:

₁₀₁²⁶⁰Cf —> ₁₀₂²⁶⁰No + _-1⁰B