Question

In: Chemistry

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

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

Solutions

Expert Solution

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


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