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Samir Lal Vig
Asked: 2022-11-04T18:49:41+05:30 In:

In the radioacitve decay
`._(Z)X^(A) rarr ._(z + 1)Y^(A) rarr ._(z – 1)P^(A – 4) rarr (z – 1)Z^(*A – 4)`
The sequence of emission is
a. `alpha, beta, gamma` b. `gamma, alpha, beta` c. `beta, alpha, gamma` c. `beta, gamma, alpha`
A. `alpha, beta gamma`
B. `beta, alpha, gamma`
C. `gamma , alpha, beta`
D. `beta, gamma, alpha`

In the radioacitve decay
`._(Z)X^(A) rarr ._(z + 1)Y^(A) rarr ._(z – 1)P^(A – 4) rarr (z – 1)Z^(*A – 4)`
The sequence of emission is
a. `alpha, beta, gamma` b. `gamma, alpha, beta` c. `beta, alpha, gamma` c. `beta, gamma, alpha`
A. `alpha, beta gamma`
B. `beta, alpha, gamma`
C. `gamma , alpha, beta`
D. `beta, gamma, alpha`
Parabola
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1 Answer

  1. Correct Answer – b
    `._(Z)^(A)X – ._(-1)^(0)e to ._(Z + 1)^(A)Y, ._(Z + 1)^(A)Y – ._(2)^(4)He to ._(Z – 1)^(A – 4)Z`,
    `._(Z – 1)^(A – 4)Z – gamma to ._(Z – 1)^(A – 4) Z**`

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