A hypothetical hydrogen-like atom consists of a proton and a negatively charged unknown particle. The ground state energy of this atom is found to be -2720 eV . Given that the first Bohr radius of a normal hydrogen atom is 0.53 and its ground state energy is -13.6 eV , what is the radius of the second Bohr orbit of this hypothetical atom?

Step-by-step solution

The energy of an electron in a Bohr orbit is directly proportional to the mass of the orbiting particle: E m . Let the mass of the unknown particle be m_x . The ratio of its mass to the electron's mass is: m_x m_e = E_x E_e = -2720 eV -13.6 eV = 200 Thus, m_x = 200 m_e . The radius of a Bohr orbit is given by r_n n^2 m . For the first orbit ( n=1 ) of the normal hydrogen atom: r₁ = 0.53 For the second orbit ( n=2 ) of the hypothetical atom, the radius is: r₂' = r₁ ( n^2 1^2 ) ( m_e m_x ) r₂' = 0.53 4 ( 1 200 ) r₂'