90 232 Th disintegrates to 82 208 Pb- How many of -particles are evolved-

_ 90 ^ 232 Th ⟶ _ 82 ^ 208 Pb +6_ 2 ^ 4 He +4_ 1 ^ 0 β #160; There are 4 β particles. ...

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Standard XII

Chemistry

Classification of Colloids - Based on Types of Particles of Dispersed Phase

90 232 Th dis...

Question

$_{90}^{232} T h$ disintegrates to $_{82}^{208} P b$ . How many of $β$ -particles are evolved?

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Solution

$_{90}^{232} T h ⟶_{82}^{208} P b + 6_{2}^{4} H e + 4_{- 1}^{0} β$
There are 4 $β$ particles.

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Similar questions

_{90} T h^{234}

disintegrates to give

_{82} P b^{206}

as the final product. How many

α

and

β - p a r t i c l e s

are emitted out during this process?

_{90}^{234} T h

disintegrates to give

_{82}^{206} P b

as the final product. The number of

α

- and

β

-particles are emitted during the process:

Q. How many

α

and

β

particles are emitted when

_{92} U^{238}

disintegrates to form lead

_{82} {P b}^{206}

_{92}^{235} U

decays with emission of

α

and

β -

particles to form ultimately

_{82}^{207} P b

. How many

α -

and

β -

particles are emitted per atom of

P b

produced?

Thorium with atomic number 92 and mass number 238 disintegrates to give Pb with atomic number 82 and mass number 206. How many alpha and beta particles are emitted during this process?

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23290Th disintegrates to 20882Pb. How many of β-particles are evolved?

23290Th⟶20882Pb+642He+40−1βThere are 4 β particles.

$_{90}^{232} T h$ disintegrates to $_{82}^{208} P b$ . How many of $β$ -particles are evolved?

$_{90}^{232} T h ⟶_{82}^{208} P b + 6_{2}^{4} H e + 4_{- 1}^{0} β$
There are 4 $β$ particles.