Home
Class 12
CHEMISTRY
The isotope ""(y)A^(x) undergoes a serie...

The isotope `""_(y)A^(x)` undergoes a series of `m alpha` and `n beta` disintegrations to form a stable isotope `""_(y-10)B^(x-32)` . The values of m and n are

A

6 and 8

B

8 and 10

C

5 and 8

D

8 and 6

Text Solution

Verified by Experts

The correct Answer is:
D
`""_(Y)A^X to ""_(Y-10)B^(X-32) + m_2 He^4 + n ""_(+1)e^0`
value of m `=(X-(X)-32)/(4) = 8`
Value of `n = Y-Y-10-2 xx 8 = 6`
Promotional Banner

Similar Questions

Explore conceptually related problems

The isotope ._(92)U^(238) successively undergoes eight alpha -decays and six beta -decays. The resulting isotope is found to be ._(92-5y)X^(238-4x) . Find the value of ratio x//y .

When the radioactive isotope _(88)Ra^(226) decays in a series by emission of three aplha (alpha) and a beta (beta) particle, the isotope X which remains undecay is

When the radioactive isotope ._(88)Ra^(228) decays in series by the emission of 3 alpha and 1 beta particle, the isotope finally formed is

In the actinium, series, the first member ._(92)U^(235) emits in succession 7 alpha particles and 4 beta particles and gets converted into a stable lead isotope lead isotope. Find the atomic number and mass number of the lead isotope so formed.

._(n)X^(m) emitted one alpha and 2beta particles, then it will become :

in (M)=x (Number of Geometrial isomers in (N)=y). The value of (y)/(x) is