The `Th_(90)^(232)` atom has successive alpha and beta decays to the end product `Pb_(82)^(208)` . The numbers of alpha and beta. Particles. Emitted i

Question

The `Th_(90)^(232)` atom has successive alpha and beta decays to the end product `Pb_(82)^(208)` . The numbers of alpha and beta. Particles. Emitted i

The `Th_(90)^(232)` atom has successive alpha and beta decays to the end product `Pb_(82)^(208)` . The numbers of alpha and beta. Particles. Emitted in the process respecitvely are-
A. 4,6
B. 4,4
C. 6,2
D. 6,4

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

A nucleus with Z =92 emits the following in a sequence: `alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(

2 Answers 5 Views

Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)`

2 Answers 4 Views

Assuming that only particles emitted from atoms during natural radioactive decay are alpha and beta particles, which of the following atoms could not

2 Answers 4 Views

A radioactive element decays by `beta-`emission. A detector racords `n`-beta particles in `2` sec and in next `2` sec it records `0.65n`-beta particle

2 Answers 5 Views

In the following radioactive decay `x^(232)rarry^(220)` how many `alpha` and `beta` particles are ejected from x to y?

2 Answers 5 Views

If a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then th

2 Answers 4 Views

If `"sin"(alpha + beta) "sin" (alpha-beta) = "sin" gamma(2"sin" beta + "sin"gamma) " where " 0 lt alpha, beta, lt pi,` then the straight line whose eq

2 Answers 8 Views

If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to

2 Answers 6 Views

If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),

2 Answers 4 Views

The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `

2 Answers 5 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
Mass number get charge by `rArr4xxn`
because atomic number charge occur only in `alpha`
decay
4n=232-208
4n=24
4n=6
`alpharArr4`
Mass number get charge by `rArr24`
New atomic number `rArr90-6xx2+P`
90-12+P=82
P=94-90=4
`betarArr4`

5 views Answered 2023-02-05 15:29