Question

What is the respective number of α and β- particles emitted in the following radioactive decay
200^X90 ----> 168^Y80 ?
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Answer 1

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=> 8 and 6 !

Explanation:

Suppose x α-particles and yβ- particles are emitted So, change in mass no. is given by

4x = 200 - 168 = 32

x = 8

and change in atomic no. is given by

2x = 8 - y = 10

Putting value of x

2 × 8 - y = 10

So, no of β- particles y = 6

No. of α-particles x = 8

Alternative

200^X90 ------> 168^Y90

200^X90 -------> ( n2 He^4 ) + m( -1β^0 ) + 168^Y80

Therefore, in this reaction

200 = 4n + 168 or n = 200-168 / 4 = 8

Also, 90 = 2n - m + 80

or m = 2n + 80 - 90

m = 2 × 80 - 90 = 6

Thus, respective number of α and β-particles will be 8 and 6 !

Answer 2

Answer:

$\bigstar$ Explanation:

Suppose x α-particles and yβ- particles are emitted So, change in mass no. is given by

4x = 200 - 168 = 32

x = 8

and change in atomic no. is given by

2x = 8 - y = 10

Putting value of x

2 × 8 - y = 10

So, no of β- particles y = 6

No. of α-particles x = 8

Alternative

200^X90 ------> 168^Y90

200^X90 -------> ( n2 He^4 ) + m( -1β^0 ) + 168^Y80

Therefore, in this reaction

200 = 4n + 168 or n = 200-168 / 4 = 8

Also, 90 = 2n - m + 80

or m = 2n + 80 - 90

m = 2 × 80 - 90 = 6

Thus, respective number of α and β-particles will be 8 and 6 !