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Rutherford's Model

In successive...

Question

In successive emission of α and β-particles, the number of α and β-particles should be emitted for the conversion of $_{92} U^{238}$ to $_{82} P b^{206}$ are:

A

7 α, 5 β

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B

6 α, 4 β

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C

4 α, 3 β

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D

8 α, 6 β

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Solution

The correct option is C $8 α, 6 β$

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Similar questions

Q. In successive emission of

α

β

- particles, the number of

α

β

- particles that should be emitted for the conversion of

_{92} U^{238}

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

Q. The number of

α

β^{-}

emitted during the radioactive decay chain starting from

\begin{matrix} 22688 \end{matrix} R a

\begin{matrix} 20682 \end{matrix} P b

Q. ( a ) Calculate number of

α

β -

particles emitted when

_{92} U^{238}

changes into radioactive

_{82} P b^{206} .

T h^{234}

disintegrates and emits

6 β -

7 α -

particles to form a stable element. Find the atomic number and
mass number of the stable product. Also identify the element:

Q. By the successive disintegration of

_{92} U^{238}

, the final product obtained is

_{82} P b^{206}

, then how many number of

α

β

particles are emitted?

Q. Match the following:

List-I Series	List-II Particles emitted
$(A)$ Thorium	$(i)$ $8 α, 5 β$
$(B)$ Neptunium	$(i i)$ $8 α, 6 β$
$(C)$ Actinium	$(i i i)$ $6 α, 4 β$
$(D)$ Uranium	$(i v)$ $7 α, 4 β$

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