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The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV. (a) What is the kinetic energy of the electron in this state? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed? |
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Answer» Solution :According to hypothesis of Bohr model. `mvr=(nh)/(2pi)` and centripetal force = coulomb force. `(mv^(2))/(r)=(e^(2))/(4p epsi_(0)r^(2)) [ :.` For H, Z=1] `:.(1)/(2) mv^(2)= (e^(2))/(8 pi epsi_(0)r)` `:.k=(e^(2))/(8 pi epsi_(0)r)` `:. K=(e^(2))/(8 pi epsi_(0)r )""...(1)` and potential energy , `U=-(e-e)/(4pi epsi_(0)r)=(-e^(2))/(4pi epsi_(0)r) ""...(2)` `:.U=-((e^(2))/(8 pi epsi_(0)r))=-2K""...(3) ` and E=K+U `=K+(-2k)` `:.E=-K ""..(4)` E = -3.4 eV depend on the conventional CHOICE in which the potential energy is ZERO at infinite distance. `E=-k` `:.-(3.4)=-K` `:.K=3.4eV` |
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