Answer
413.7k+ views
Hint: We know that the density of any nucleus is constant. So the volume of the nucleus will be proportional to the mass number of the nucleus. From the volume we can get the radius of the nucleus. According to Bohr, there is a formula to find out the radius of Bohr's orbit and this is only applicable for hydrogen like species.
Formula used:
$R = 0.53 \times \dfrac{{{n^2}}}{Z}{A^ \circ }$
Complete answer:
Hydrogen-like species in the sense of atoms which have only one electron. Normally hydrogen has one electron only. If we remove one electron from helium then it also has one electron. Similarly removing two electrons from lithium converts lithium into hydrogen like species. In this manner there is a formula given by bohr to find out the radius of bohr orbit which is valid for all the hydrogen like atoms. That formula is
$R = 0.53 \times \dfrac{{{n^2}}}{Z}{A^ \circ }$
Where ‘n’ is the principal quantum number and Z is the atomic number. Above radius is measured in angstroms. That means order of ${10^{ - 10}}m$.
Now coming to the radius of the nucleus, it will be proportional to mass number. But whatever may be the mass number, only value changes but order will remain the same. It will be of order ${10^{ - 14}}m$.
Hence option A is correct.
Note:
In between the particles of nucleus i.e nucleons, strong nuclear force will exist. That is the strongest fundamental force and between the electron in bohr’s orbit and nucleus, an electromagnetic force will be existing and it is the second strongest force in nature. Strong nuclear force is short ranged force and electromagnetic force long ranged force.
Formula used:
$R = 0.53 \times \dfrac{{{n^2}}}{Z}{A^ \circ }$
Complete answer:
Hydrogen-like species in the sense of atoms which have only one electron. Normally hydrogen has one electron only. If we remove one electron from helium then it also has one electron. Similarly removing two electrons from lithium converts lithium into hydrogen like species. In this manner there is a formula given by bohr to find out the radius of bohr orbit which is valid for all the hydrogen like atoms. That formula is
$R = 0.53 \times \dfrac{{{n^2}}}{Z}{A^ \circ }$
Where ‘n’ is the principal quantum number and Z is the atomic number. Above radius is measured in angstroms. That means order of ${10^{ - 10}}m$.
Now coming to the radius of the nucleus, it will be proportional to mass number. But whatever may be the mass number, only value changes but order will remain the same. It will be of order ${10^{ - 14}}m$.
Hence option A is correct.
Note:
In between the particles of nucleus i.e nucleons, strong nuclear force will exist. That is the strongest fundamental force and between the electron in bohr’s orbit and nucleus, an electromagnetic force will be existing and it is the second strongest force in nature. Strong nuclear force is short ranged force and electromagnetic force long ranged force.
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