Home
Class 12
CHEMISTRY
One of naturally occurring radioactive d...

One of naturally occurring radioactive decay series begins with `_(90)^(232)Th` and ends with a stable `_(82)^(208)Pb`.
How much energy in MeV is released in the complete chain?

Text Solution

Verified by Experts

The correct Answer is:
`_(90)^(232)Th rarr `_(82)^(208)Pb+6`_(2)^(4)He+4beta^(-)```
Energy released is `Q` value
`Q=[m(`^(232)Th)-m(`^(208)Pb)-6m(`^(4)He)]c^(2)````
(the mass of `4e^(-)` are included in daughters)
`=[232.03805 u-207.97664 u-6xx4.00260 u]xx931.5 Me V u^(-1)=`
`=(0.04581u)xx(931.5 MeV)=42.67 MeV`
Promotional Banner

Similar Questions

Explore conceptually related problems

One of naturally occurring radioactive decay series begins with _(90)^(232)Th and ends with a stable _(82)^(208)Pb . How many beta (beta) decays are there in this series? Show by calculate.

._(90)^(232)Th disintegrates to ._(82)^(208)Pb . How many of beta -particle are evolved?

How may alpha- and beta- particles will be emitted when ._(90)Th^(232) changes into ._(82)Pb^(208) ?

After a series of alpha and beta decays, ._94Pu^(239) becomes ._82Pb^(207) . How many alpha and beta particles are emitted in the complete decay process?

A nuclear reaction that uses one nucleus of ""_(92)^(236)U generates 170 Mev. How much energy is released when 5.0 kg of this isotope is udeD?

A radioactive disintegration of ""_(90)Th^(232) yields ""_(82)Pb^(208) in the end . The number of alpha and beta -particle emitted will be

The number of alpha and beta - emitted during the radioactive decay chain starting from ._(226)^(88) Ra and ending at ._(206)^(82) Pb is

._(90)Th^(232) to ._(82)Pb^(208) . The number of alpha and beta-"particles" emitted during the above reaction is