Question

Part A

24195Am→23793Np+_____ Express your answer in isotopic notation.

Part B

_____+0−1e→12655Cs Express your answer in isotopic notation.

Part C

23793Np→_____+42He Express your answer in isotopic notation.

Part D

21885At→_____+42He Express your answer in isotopic notation.

Answer 1

A)

Let the product be X

Balance mass number on both sides

mass of Am = mass of Np + mass of X

1*241 = 1*237 + M

M = 4

Balance atomic number on both sides

atomic of Am = atomic of Np + atomic of X

1*95 = 1*93 + A

A = 2

Atomic number 2 is for element He

So, the product is

4-He

⁴₂He

B)

Let the reactant be X

Balance mass number on both sides

mass of e + mass of X = mass of Cs

1*0 + M = 1*126

M = 126

Balance atomic number on both sides

atomic of e + atomic of X = atomic of Cs

1*-1 + A = 1*55

A = 56

Atomic number 56 is for element Ba

So, the reactant is

126-Ba

¹²⁶₅₆Ba

C)

Let the product be X

Balance mass number on both sides

mass of Np = mass of He + mass of X

1*237 = 1*4 + M

M = 233

Balance atomic number on both sides

atomic of Np = atomic of He + atomic of X

1*93 = 1*2 + A

A = 91

Atomic number 91 is for element Pa

So, the product is

233-Pa

²³³₉₁Pa

D)

Let the product be X

Balance mass number on both sides

mass of At = mass of He + mass of X

1*218 = 1*4 + M

M = 214

Balance atomic number on both sides

atomic of At = atomic of He + atomic of X

1*85 = 1*2 + A

A = 83

Atomic number 83 is for element Bi

So, the product is

214-Bi

Answer: ²¹⁴₈₃Bi