147508 What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus.?

147509 If $M_{O}$ is the mass of oxygen isotope ${ }_{8} O^{17}, M_{p}$ and $M_{n}$ are the masses of proton and a neutron, respectively, the nuclear binding energy of the isotope is

147510 The ratio of the binding energies of three nuclei is $1: 4: 9$, If the ratio of their nuclear radii is 1 $: 2: 3$, the nuclei in the decreasing order of the stability can be arranged as

147511 Using the following data
Mass of hydrogen atom $=1.00783 \mathrm{u}$
Mass of neutron $=1.00867 \mathrm{u}$
Mass of nitrogen atom $\left({ }_{7} \mathrm{~N}^{14}\right)=14.00307 u$
The calculated value of the binding energy of the nucleus of the nitrogen atom $\left.{ }_{7} \mathbf{N}^{14}\right)$ is close to

147508 What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus.?

147509 If $M_{O}$ is the mass of oxygen isotope ${ }_{8} O^{17}, M_{p}$ and $M_{n}$ are the masses of proton and a neutron, respectively, the nuclear binding energy of the isotope is

147510 The ratio of the binding energies of three nuclei is $1: 4: 9$, If the ratio of their nuclear radii is 1 $: 2: 3$, the nuclei in the decreasing order of the stability can be arranged as

147511 Using the following data
Mass of hydrogen atom $=1.00783 \mathrm{u}$
Mass of neutron $=1.00867 \mathrm{u}$
Mass of nitrogen atom $\left({ }_{7} \mathrm{~N}^{14}\right)=14.00307 u$
The calculated value of the binding energy of the nucleus of the nitrogen atom $\left.{ }_{7} \mathbf{N}^{14}\right)$ is close to

147508 What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus.?

147509 If $M_{O}$ is the mass of oxygen isotope ${ }_{8} O^{17}, M_{p}$ and $M_{n}$ are the masses of proton and a neutron, respectively, the nuclear binding energy of the isotope is

147510 The ratio of the binding energies of three nuclei is $1: 4: 9$, If the ratio of their nuclear radii is 1 $: 2: 3$, the nuclei in the decreasing order of the stability can be arranged as

147511 Using the following data
Mass of hydrogen atom $=1.00783 \mathrm{u}$
Mass of neutron $=1.00867 \mathrm{u}$
Mass of nitrogen atom $\left({ }_{7} \mathrm{~N}^{14}\right)=14.00307 u$
The calculated value of the binding energy of the nucleus of the nitrogen atom $\left.{ }_{7} \mathbf{N}^{14}\right)$ is close to

147508 What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus.?

147509 If $M_{O}$ is the mass of oxygen isotope ${ }_{8} O^{17}, M_{p}$ and $M_{n}$ are the masses of proton and a neutron, respectively, the nuclear binding energy of the isotope is

147510 The ratio of the binding energies of three nuclei is $1: 4: 9$, If the ratio of their nuclear radii is 1 $: 2: 3$, the nuclei in the decreasing order of the stability can be arranged as

147511 Using the following data
Mass of hydrogen atom $=1.00783 \mathrm{u}$
Mass of neutron $=1.00867 \mathrm{u}$
Mass of nitrogen atom $\left({ }_{7} \mathrm{~N}^{14}\right)=14.00307 u$
The calculated value of the binding energy of the nucleus of the nitrogen atom $\left.{ }_{7} \mathbf{N}^{14}\right)$ is close to