MyQuestionIcon

8

You visited us 8 times! Enjoying our articles? Unlock Full Access!

Beta Decay

The isotope ...

Question

The isotope $_{92}^{238} U$ decays successively to form $_{90}^{234} T h,$ $_{91}^{234} P a,$ $_{92}^{234} U,$ $_{90}^{230}$ $T h$ and $_{88}^{226} R a$ . The radiations emitted in these five steps are

A

α, α, α, β, β

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

α, α, β, β, α

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

α, β, β, α, α

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

β, β, α, α, α

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $α, β, β, α, α$
$_{92}^{238} U \to_{90}^{234} T h \to_{91}^{234} P a \to_{92}^{234} U \to_{90}^{230} T h \to_{88}^{226} R a$
A -4 0 0 -4 -4
Z -2 +1 +1 -2 -2

$α β β α α$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A nucleus with

Z = 92

, emits the following in a sequence,

α, β^{-}, β^{-}, α, α, α, α, α, β^{-},

β^{-}, α, β^{+}, β^{+}, α

Z

of the resulting nucleus is ______.

Q. A nucleus with

Z = 92

emits the following in a sequence

α, α, β^{-}, β^{-}, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α

Z

of the resulting nucleus is

Q. A nucleus with

Z = 92

emits the following in a sequence:

α, α, β^{-}, β^{-}, α, α, α, α; β^{-}, β^{-}, α, β^{+}, β^{+}, α

Z

of the resulting nucleus is

Q. A nucleus with Z = 92 emits the following in a sequence:

α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α,

The Z of the resultant nucleus is [AIEEE-2003]

A nucleus with Z = 92 emits the following in a sequence: $α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α,$ The Z of the resultant nucleus is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Beta decay

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard X Physics

Join BYJU'S Learning Program

CrossIcon