Question

Write equations for the following reactions. Alpha decay of Th 290, U 238, Pu 244

Answer 1

1) for Th-290

Let the product be X

Balance mass number on both sides

mass of Th = mass of He + mass of X

1*290 = 1*4 + M

M = 286

Balance atomic number on both sides

atomic of Th = atomic of He + atomic of X

1*90 = 1*2 + A

A = 88

Atomic number 88 is for element Ra

So, the product is

286-Ra

Answer:

₉₀²⁹⁰Th —> ₈₈²⁸⁶Ra + ₂⁴He

2) for U-238

Let the product be X

Balance mass number on both sides

mass of U = mass of He + mass of X

1*238 = 1*4 + M

M = 234

Balance atomic number on both sides

atomic of U = atomic of He + atomic of X

1*92 = 1*2 + A

A = 90

Atomic number 90 is for element Th

So, the product is

234-Th

Answer:

₉₂²³⁸U —> ₉₀²³⁴Th + ₂⁴He

3) for Pu-244

Let the product be X

Balance mass number on both sides

mass of Pu = mass of He + mass of X

1*244 = 1*4 + M

M = 240

Balance atomic number on both sides

atomic of Pu = atomic of He + atomic of X

1*94 = 1*2 + A

A = 92

Atomic number 92 is for element U

So, the product is

240-U

Answer:

₉₄²⁴⁴Pu —> ₉₂²⁴⁰U + ₂⁴He