[" Assuming the radius of the nucleus to be given as "R=R_(0)A^(1/3)" ,where "],[R_(0)=1.1times10^(-15)m" and "A" is the mass number of nucleus.Mass of "],[" nucleon is "m=1.66times10^(-27)kg" .Then mark the correct options."]

If A is mass number of a nucleus of Radius R, then

The radius of a nucleus is given by r_(0) A^(1//3) where r_(0) = 1.3xx 10^(-15) m and A is the mass number of the nucleus, the Lead nucleus has A = 206. the electrostatic force between two protons in this nucleus is approximately

Atomic nucleus is central core of every atom in which the whole of positive charge and almost entire mass of atom is concentrated. It is a tiny sphere of a radius R is given by R=R_(o)A^(1//3) , where R_(o)=1.4xx10^(-15)m , a constant and A the mass number of nucleus The radius of the nucleus of mass number 125 is

The radius of the nucleus of the atom with A=216 is (take R_(0)=1.3 fm)

From the relation R=R_(0)A^(1//3) , where 'R_(0)' is a constant and 'A' is the mass number of a nucleus, show that the nuclear matter density is nearly constant.

Radius of a nucleus is given by the relation R = R_(0)A^(1//3) where R_(0) = 1.3 xx 10^(-15) m and A is mass number. For a nucleon inside a nucleus, de-Broglie wavelength is given by diameter of the nucleus. Average kinetic energy of a nucleon in the Te^(128) nucleus based on above information will be :

Radius of a nucleus is given by the relation R = R_(0)A^(1//3) where R_(0) = 1.3 xx 10^(-15) m and A is mass number. For a nucleon inside a nucleus, de-Broglie wavelength is given by diameter of the nucleus. Average kinetic energy of a nucleon in the Te^(128) nucleus based on above information will be :

Atomic nucleus is central core of every atom in which the whole of positive charge and almost entire mass of atom is concentrated. It is a tiny sphere of a radius R is given by R=R_(o)A^(1//3) , where R_(o)=1.4xx10^(-15)m , a constant and A the mass number of nucleus On increasing the value of 'A' the density of the nucleus

Calculate radius of nucleus oxygen.(Given: R_0 =1.1 xx 10^-15 m)