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Obtain approximately the ratio of the nu...

Obtain approximately the ratio of the nuclear radii of the gold isotope `"_79^197Au` and the silver isotope `"_47^107 Ag`.

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In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be 100:1. The mean lives of the two isotopes are 4xx10^9 years and 2xx10^9 years, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal propotional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is 1.02:1.

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive "_6^14C present with the stable carbon isotope "_6^12C . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of "_6^14C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of "_6^14C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

An alpha particle (_2He^4) is moving directly towards a stationary gold nucleus (_(79)Au^(197)) .The alpha -particle and the gold nucleus may be considered to be solid spheres with the charge and mass concentrated at the centre of each sphere.When the two spheres are just toching,the sepaation of their centres is 9.6 xx 10^(-15) m . the alpha particle and the gold nucleus may be assumed to be an isolatedd system.Calculate for the alpha particle just in contact with the gold nucleus, its electric potential energy.

A radioactive isotope X with half-life of 693xx10^(9) years decay to Y which is stable. A sample of rock from of the moon was found to contain both the elements X Y in the mole ratio 1:7 . What is the age of the rock ? a) 2.079xx10^(12) years b) 1.94xx10^(10) years c) 1.33xx10^(9) years d) 10^(10) years

._(83)^(214)Bi decays to A by alpha -emission . A then decays to B by beta emission , which further decays to C by another beta emission . Element C decays to D by still another beta emission , and D deacays by alpha -emission to form a stable isotope E. What is element E? a) _(81)^(207)T1 b) _(80)^(206)Hg c) _(79)^(206)Au d) _(82)^(206)Pb

Silver is electrodeposited on a metallic vessel of total surface area 800cm^2 by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of silver deposited given density is 10.47g/cc. (At. Wt. of Ag=107.92).

The isotopes .^(238)U and .^(235)U occur in nature in the mass ratio 140 : 1. it is assumed that intially they were found in equal mass. If half life (t_(1//2)) of .^(238)U=4.5xx10^(9) and t_(1//2) of .^(235)U 7.13 x10^(8) years respectively then the age of earth is (log 7=0.846,log2=0.3)