Find the atomic number and mass number of the last member in the following series:
(a) $_{88}^{226} R a - α - ---- \to R n - α - ---- \to R a A - α - ---- \to R a B - β - ---- \to R a C$
(b) $_{Z}^{M} A - α - ---- \to B - β - ---- \to C - β - ---- \to D - α - ---- \to E$

Open in App

Solution

By Emission of $α$ particle, atomic number decreases by 2 units and mass number by 4 units.
By emission of $β$ particle atomic number increases by 1 unit.
(a)3 $α$ particles are emitted by Ra
$∴$ Mass number reduces by (3 $\times$ 4)= 12 units
Atomic number reduces by 3 $\times$ 2 =6 units
1 $β$ particle is emitted further
$∴$ Atomic number of Ra increases by 1 unit.
In total Mass number of Ra reduces by 12 units and atomic number by 5 units.
$∴_{88}^{226} R a {- 1 β - - \to - 3 α}_{83}^{214} R a C$
(b) Atomic number reduces by (2x2)=4 units a two alpha particles are emitted by A.
2 $β$ particles are emitted so atomic number increases by 2 units.
In total atomic number decreases by (2x4)= 8 units by emission of 2 $α$ particles.
$β$ emission has no effect on mass number.
$∴_{Z}^{M} A - 2 β - - \to - 2 α_{Z - 2}^{M - 8} E$

Suggest Corrections

Similar questions

Q. Radioactive disintegration of

_{88}^{226} R a

takes place in the following manner into

R a C

R a - α - ---- \to R n - α - ---- \to R a A - α - ---- \to R a B - β - ---- \to R a C

Determine mass number and atomic number of

R a C

Q. To which group of the periodic table does the last member of the following series belong?
(a)

_{92}^{239} U {- β - ---- \to}_{93}^{239} N p {- β - ---- \to}_{94}^{239} P u

(b)

_{55}^{140} C s {- β - ---- \to}_{56}^{140} B a {- β - ---- \to}_{57}^{140} L a {- β - ---- \to}_{58}^{140} C e

(c)

_{88}^{226} R a {- α - ---- \to}_{86}^{222} R n {- α - ---- \to}_{84}^{218} P o

Q. What is the atomic number of last member of the seventh period of the extended from of periodic table?

There are two elements A 26 mass number and 13 atomic number and B 26 mass number and 14 atomic number . find the number of sub atomic particles in each of these elements.what is the relationship between the two?

Q. A nucleus undergoes a series of decay according to the scheme

A α \to B β \to C α \to D γ \to E

. Atomic number and mass numbers of E are 69 and 172

Join BYJU'S Learning Program

Find the atomic number and mass number of the last member in the following series:(a) 22688Ra−α−−−−−→Rn−α−−−−−→RaA−α−−−−−→RaB−β−−−−−→RaC(b) MZA−α−−−−−→B−β−−−−−→C−β−−−−−→D−α−−−−−→E

Find the atomic number and mass number of the last member in the following series:
(a) $_{88}^{226} R a - α - ---- \to R n - α - ---- \to R a A - α - ---- \to R a B - β - ---- \to R a C$
(b) $_{Z}^{M} A - α - ---- \to B - β - ---- \to C - β - ---- \to D - α - ---- \to E$