The mass of $$_{7}N^{15}$$ is $$15.0011\ amu$$, mass of $$_{8}O^{16}$$ is $$15.99492\ amu$$ and $$m_{P} = 1.00783\ amu$$. Determine binding energy of last proton of $$_{8}O^{16}$$.
$$M(_{8}O_{16}) = M(_{7}N^{15}) + 1m_{P}$$
binding energy of last proton
$$= M(N^{15}) + m_{P} - M(_{1}O^{16})$$
$$= 15.00011 + 1.00783 - 15.99492$$
$$= 0.01302\ amu = 12.13\ MeV$$.
Assume that a neutron breaks into a proton and an electron. The energy released during this process is (Mass of neutron $$=1.6725\times 10^{-27} kg$$, Mass of proton $$=1.6725\times 10^{-27} kg$$ ,Mass of electron $$=9\times 10^{-31} kg$$)
The binding energy per nucleon for the parent nucleus is $$\mathrm{E}_{1}$$ and that for the daughter nuclei is $$\mathrm{E}_{2}$$.Then
Find $$BE$$ per nucleon of $$^{56}Fe$$ where $$m(^{56}Fe) = 55.936u\ m_{n} = 1.00727u, m_{p} = 1.007274\ u$$.
Find the binding energy of a H-atom in the state n=2
The atomic mass of $$B^{10}$$ is $$10.811\ amu$$. The binding energy of $$B^{10}$$ nucleus is [ Given: The mass of electron is $$0.0005498\ amu$$, the mass of proton is $$m_p=1.007276\ amu$$ and the mass of neutron is $$m_n=1.008665\ amu$$]:
The binding energy of $$Na^{23}$$ is [ Given : Atomic mass of $$22.9898\ amu$$ and that of $$^1 H_1$$ is $$1.00783\ amu$$]:
What is the binding energy per nucleon of $$_{6}C^{12}$$ nucleus? Given : mass of $$C^{12}(m_c)=12.000\ u$$ mass of proton $$9M_p)=1.0078\ u$$ mass of neutron $$(m_n)=1.0087\ u$$ and $$1\ amu=931.4\ MeV$$
When number of nucleons in nuclei increases, the binding energy per nucleon:
The binding energy of an electron in the ground state of $$He\ $$atom is $$E_0=24.6 eV.$$ The energy required to remove both the electrons from the atom is
Find the binding energy of valence electron in the ground state of a $$Li$$ atom if the wavelength of the sharp series is known to be $$\lambda_{1}$$ $$= 813 nm$$ and the short wave cutoff wavelength $$\lambda_{2}$$, $$= 350 nm$$