When 92U238 decays it emits an - - particle- The new nuclide in turn emits a - - particle to give another nuclide X- The mass number and atomic number of X are respectively -

The correct option is A 234 and 91 _92U^238→ _zX^A + _2He^4 + _ 1e^0 Equating mass number of both sides238 = A + 4 + 0 ∴ A = 238 4 =234Equating atom ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

pH the Power of H

When 92U238...

Question

When $_{92} U^{238}$ decays it emits an $α$ - particle. The new nuclide in turn emits a $β$ - particle to give another nuclide X. The mass number and atomic number of X are respectively :

234 and 91

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

234 and 96

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

232 and 88

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

234 and 88

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

Open in App

Solution

The correct option is A 234 and 91
$_{92} U^{238} \to_{z} X^{A} +_{2} H e^{4} +_{- 1} e^{0}$
Equating mass number of both sides
238 = A + 4 + 0
$∴$ A = 238 - 4 =234
Equating atomic number of both sides
92 = Z + 2 - 1
$∴$ Z = 92 - 1 = 91

Suggest Corrections

Similar questions

Q. A nuclide (X) of mass number 233 emits six

α

- particles and same number of

β

- particles. The mass number of the daughter nuclide is

Q. The atomic number (

A

) and mass number (

M

) of the nuclide formed when three alpha

α

and two

β

particles are emitted from

238_{92}^{238} U

Q. In the radioactive decay process of uranium the initial nuclide is

_{92} U^{238}

and the final nuclide is

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

. When uranium nucleus decays to lead, then the number of

α

-particles and

β

-particles emitted will respectively be :

Q. The nuclide

X

undergoes

α

-decay and another nuclide

Y

β^{-}

-decay. Which of the following statements are correct?
1. The

β

-particles emitted by

Y

may have widely different speeds.
2. The

α

-particles emitted by

X

may have widely different speeds.
3. The

α

-particles emitted by

X

will have almost same speed.
4. The

β

-particles emitted by

Y

will have the same speed.

Q. The uranium nucleus

_{92} U^{238}

emits an

α

-particle and resulting nucleus emits one

β

-particle. The atomic number and mass number of the final nucleus will be respectively

Join BYJU'S Learning Program

When 92U238 decays it emits an α - particle. The new nuclide in turn emits a β - particle to give another nuclide X. The mass number and atomic number of X are respectively :

The correct option is A 234 and 9192U238→zXA+2He4+−1e0Equating mass number of both sides238 = A + 4 + 0∴ A = 238 - 4 =234Equating atomic number of both sides92 = Z + 2 - 1∴ Z = 92 - 1 = 91

When $_{92} U^{238}$ decays it emits an $α$ - particle. The new nuclide in turn emits a $β$ - particle to give another nuclide X. The mass number and atomic number of X are respectively :

The correct option is A 234 and 91
$_{92} U^{238} \to_{z} X^{A} +_{2} H e^{4} +_{- 1} e^{0}$
Equating mass number of both sides
238 = A + 4 + 0
$∴$ A = 238 - 4 =234
Equating atomic number of both sides
92 = Z + 2 - 1
$∴$ Z = 92 - 1 = 91