Home
Class 12
CHEMISTRY
1 g of .(79)Au^(198) (t(1//2) = 65 hr) d...

`1 g` of `._(79)Au^(198) (t_(1//2) = 65 hr)` decays by `beta`-emission to produce stable `Hg`.
a. Write nuclear reaction for process.
b. How much `Hg` will be present after 260 hr.

Text Solution

Verified by Experts

a. `._(79)Au^(198) rarr ._(80)Hg^(198) ._(-1)e^(0)`
b. `t_(1//2) = 65 hr`
`T = 260 hr`
`:' T = t_(1//2) xx n`
`:.` Number of halves `(n) = (260)/(65) = 4`
Now, amount left undcayed `(N) = (N_(0))/(2^(4)) = (1)/(2^(4)) = (1)/(16) g`
`:.` Amount of `Au` decayed `= (15)/(16) g`
`:' g Au` gives `198 g Hg`.
`:. (15)/(16) Au` gives `(15)/(6) Hg`.
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY|Exercise Solved Example|18 Videos

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY|Exercise Ex6.1 Objective|15 Videos

  • NCERT BASED EXERCISE

    CENGAGE CHEMISTRY|Exercise Nuclear Chemistry (NCERT Exercise)|29 Videos

  • ORGANIC COMPOUNDS WITH FUNCTIONAL GROUP

    CENGAGE CHEMISTRY|Exercise Archives Analytical And Descriptive|24 Videos

Similar Questions

Explore conceptually related problems

One gram of ._(79)Au^(198) (t_(1//2) = 65hr) decays by beta- emission to produce stable Hg . (a) Write nuclear reaction for process. (b) How much Hg will be present after 260 hr ?

1 g of ._(79)Au^(198)(t_(1//2)=65h) gives stable mercury by beta- emission. What amount of mercury will left 260 h?

._(84)Po^(218) (t_(1//2) = 3.05 min) decays to ._(82)Pb^(214) (t_(1//2) = 2.68 min) by alpha emisison while Pb^(214) is beta -emitter. In an experiment starting with 1 g atom of pure Po^(218) , how much time would be required for the concentration of Pb^(214) to reach maximum?

A radioisotope ._(Z)A^(m) (t_(1//2) = 10 days) decays to give ._(z - 6)B^(m - 12) stable atom along with alpha -particles. If m g of A are taken and kept in a sealed tube, how much He will accumulate in 20 days at STP .

Out of300g substance [decomposes as per 1st order]. How much (nearly)will remain after 18 hr? (t_(1//2) = 3hr)