How many alpha and beta particles emit when uranium nucleus
238. 206
U. To. Pb. ?
92. 82
238. 206
U. To. Pb. ?
92. 82
Answer: 8 alpha decays which release 8 alpha particles
Explanation: U238→Pb206
(This is not a direct decay but is a process of many alpha decays and beta decay).
An alpha particle has 2neutrons and 2protons(A helium nucleus)
Thus the combined mass is 4
You only need how many alpha decay it undergoes . This can be determined by an equation like this
206=238−(4x)
Solve for x
238−4x−238=206−238(Subtract 238 from both sides )
−4x=−32
Thus−x=−8
and x=8.
This means that this process undergoes 8 alpha or α decays which means 8nuclei of He have been emitted
Alpha decays
238U→234Th
234Th→230Th
230Th→226Ra
226Ra→222Rn
222Rn→218Po
218Po→214Pb
214Pb→210Pb
210Pb→206Pb
But when 8 alpha particles are emitted the atomic number will be 92-(8*2)=76.
Here atomic number is 82.Therefore Beta particles emitted = 82-76= 6
8 alpha decays which release 8 alpha particles
Explanation:U238→Pb206
(This is not a direct decay but is a process of many alpha decays and beta decay).
An alpha particle has 2neutrons and 2protons(A helium nucleus)
Thus the combined mass is 4
You only need how many alpha decay it undergoes . This can be determined by an equation like this
206=238−(4x)
Solve for x
238−4x−238=206−238(Subtract 238 from both sides )
−4x=−32
Thus−x=−8
and x=8.
This means that this process undergoes 8 alpha or α decays which means 8nuclei of He have been emitted
Alpha decays
238U→234Th
234Th→230Th
230Th→226Ra
226Ra→222Rn
222Rn→218Po
218Po→214Pb
214Pb→210Pb
210Pb→206Pb
But when 8 alpha particles are emitted the atomic number will be 92-(8*2)=76.
Here atomic number is 82.Therefore Beta particles emitted = 82-76= 6
