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.

A

0.93 g

B

0.85 g

C

1 g

D

0.79 g

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

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 .

._(84)^(218)Po(t_(1//2)=183 sec) decay to ._(82)Pb(t_(1//2)=161 sec) by alpha -emission, while Pb^(214) is a beta -emitter. In an experiment starting with 1 mole of pure Po^(218) , how many time would be required for the number of nuclei of ._(82)^(214)Pb to reach maximum ?

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs., find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present. [Given log_e3=1. 0986 ,\ e^(2. 1972)=9 ]

Gold ._(79)^(198)Au undergoes beta^(-) decay to an excited state of ._(80)^(198)Hg . If the excited state decays by emission of a gamma -photon with energy 0.412 MeV , the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the ._(80)^(198)Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968255 u for ._(79)^(198)Hg ).

83Bi^(2H)(t 1/2= 130 sec) decays to 81Tl^(207) by alpha -emission. In an experiment starting with 5 moles of 83Bi^(211) . how much pressure would be developed in a 350 L closed vessel at 25 C after 760 sec? [Antilog (1.759) = 57 41]

The radionuclide .^(56)Mn is being produced in a cyclontron at a constant rate P by bombarding a manganese target with deutrons. .^(56)Mn has a half-life of 2.5 h and the target contains large numbers of only the stable manganese isotopes .^(56)Mn . The reaction that produces .^(56)Mn is .^(56)Mn +d rarr .^(56)Mn +p After being bombarded for a long time, the activity of .^(56)Mn becomes constant, equal to 13.86 xx 10^(10) s^(-1) . (Use 1 n2=0.693 , Avagardo number =6 xx 10^(2) , atomic weight of .^(56)Mn=56 g mol^(-1) ). After a long time bombardment, number .^(56)Mn nuclei present in the target depends upon. (i) number of Mn nuclei present at the start of the process. (ii) half life of Mn. (iii) constant rate of production P

For the 1^("st") order reacton: A(g)rarr 2B(g)+C(s) , the t_((1)/(2))="24 min" . The reaction is carried out taking certain mass of A enclosed in a vessel in which it exerts a pressure of 400 mm Hg. The pressure of the reaction mixture in mm Hg after expiry of 48 min will be

A parallel - plate capacitor having plate area 20cm^2 and seperation between the plates 1.00 mm is connected to a battery of 12.0V . The plates are pulled apart to increase the separation to 2.0mm . (a ) calculate the charge flown through the circuit during the process . (b ) How much energy is absorbed by the battery during the process ? (c ) calculate the stored energy in the electric field before and after the process . (d ) Using the expression for the force between the plates , find the work done by the person polling the plates apart . (e ) Show and justify that no heat is produced during this transfer of charge as the separation is increased.