The energy of an electron in a hydrogen atom is described by \(E_n = -\frac{13.6}{n^2} \, \text{eV}\), where \(E_n\) is the energy and \(n\) is the principal quantum number.
Given an electron energy of -0.544 eV, we solve for \(n\) using the equation: \(-\frac{13.6}{n^2} = -0.544\).
Simplifying, we get: \(\frac{13.6}{n^2} = 0.544\).
Rearranging to solve for \(n^2\): \(n^2 = \frac{13.6}{0.544}\).
Calculating the value: \(n^2 = 25\).
Taking the square root yields: \(n = 5\).
Therefore, the quantum number corresponding to an energy of -0.544 eV is 5.
Assuming the experimental mass of \( {}^{12}_{6}\text{C} \) as 12 u, the mass defect of \( {}^{12}_{6}\text{C} \) atom is____MeV/\( c^2 \).
(Mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/\( c^2 \))
The binding energy per nucleon of \(^{209} \text{Bi}\) is _______ MeV. \[ \text{Take } m(^{209} \text{Bi}) = 208.98038 \, \text{u}, \, m_p = 1.007825 \, \text{u}, \, m_n = 1.008665 \, \text{u}, \, 1 \, \text{u} = 931 \, \text{MeV}/c^2. \]