If 90Th228 disintegrates to 83Bi212- then the number of - and - particles emitted are -

The correct option is B 4α and 1β #160; ...

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Byju's Answer

Standard VII

Chemistry

Sub-Atomic Particles

If 90Th228 ...

Question

If $_{90} {T h}^{228}$ disintegrates to $_{83} {B i}^{212}$ , then the number of $α$ and $β$ particles emitted are :

4 α

only

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7 β

only

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4 α

and

7 β

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4 α

and

1 β

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Solution

The correct option is B $4 α$ and $1 β$

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Similar questions

Q. When

_{90} T h^{228}

transforms to

_{83} B i^{212}

, then the number of the emitted

α

-and

β

-particles is, respectively

Q. What are the

α

and

β

particles emitted from

_{90} {T h}^{228} \to_{83} {B i}^{212}

Q. When

{_{90} T}^{228}

transforms to

{_{83} B i}^{212}

, then the number to the emitted

α

and

β

particles is, respectively:

Q. If α =

\tan^{- 1} (\tan \frac{5 π}{4}) and β = \tan^{- 1} (- \tan \frac{2 π}{3})

, then
(a) 4 α = 3 β
(b) 3 α = 4 β
(c) α − β =

\frac{7 π}{12}

(d) none of these

Q. If

α

and

β

are the roots of the equation

2 x^{2} - 7 x + 8 = 0

, then the equation whose roots are

(3 α - 4 β

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3 β - 4 α

) is

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If 90Th228 disintegrates to 83Bi212, then the number of α and β particles emitted are :

The correct option is B 4α and 1β

If $_{90} {T h}^{228}$ disintegrates to $_{83} {B i}^{212}$ , then the number of $α$ and $β$ particles emitted are :

The correct option is B $4 α$ and $1 β$