The energy of an electron in the \( n \)-th Bohr orbit is given by the formula: \[ E_n = \frac{-13.6 \, \text{eV}}{n^2}. \] For the first orbit (\( n = 1 \)): \[ E_1 = \frac{-13.6}{1^2} = -13.6 \, \text{eV}. \] For the third orbit (\( n = 3 \)): \[ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} = -1.51 \, \text{eV}. \] Hence, the energy in the third orbit is \( \frac{1}{9} \) of the energy in the first orbit.
Two positively charged particles \(m_1\) and \(m_2\) have been accelerated across the same potential difference of 200 keV. Given mass of \(m_1 = 1 \,\text{amu}\) and \(m_2 = 4 \,\text{amu}\). The de Broglie wavelength of \(m_1\) will be \(x\) times that of \(m_2\). The value of \(x\) is _______ (nearest integer). 