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Suppose the energy of a hydrogen- like atom is given as E_(n) = (-54.4)/(n^(2))eVwhere ninN. Calculate the following: (a) Sketch the energy levels for this atom and compute its atomic number. (b) If the atom is in ground state, compute its first excitation potential and also its ionization potential. (c) When a photon with energy 42 eV and another photon with energy 56 eV are made to collide with this atom, does this atom absorb these photons? (d) Determine the radius of its first Bohr orbit. (e) Calculate the kinetic and potential energies in the ground state. |
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Answer» Solution :(a) Given that `E_(n) = -(54.4)/(n^(2)) eV` For n = 1 , the ground state energy `E_(1) = -54.4 eV` and for n = 2, `E_(2) = -13.6 eV`. Similarly, `E_(3) = -6.04 eV, E_(4) = -3.4 eV` and so on . For large value of principal quantum number - that is, `n = infty`, we get `E_(infty) = 0 eV`.  (b) For a hydrogen like ATOM, ground state energy is `E_(1) = -(13.6)/(n^(2))Z^(2)eV` where Z is the atomic number. Hence, compairing this energy with given energy , we get , `-13.6 Z^(2) = -54.4 Rightarrow Z= pm2`. Since, atomic number cannot be negative number, Z =2. (c) The first excitation energy is `E_(1) = E_(2) - E_(1) = - 13.6 eV - (-54.4 eV) = 40.8 eV` Hence, the first excitation potential is `V_(1) = (1)/(e) E_(1) = ((40.8eV))/(e) = 40.8` volt The first ionization energy is `E_(ionization) = E_(infty) - E_(1) = 0 -(-54.4 eV)` `= 54.4 eV` Hence, the first ionization potential is `V_(ionization) = (1)/(e)E_(ionization) = ((54.4 eV))/(e)` `= 54.4` volt (d) Consider two photons to be A and B. Given that photon A with energy 42 eV andphoton B with energy 51 eV. From Bohr assumption, difference in energy levels is equal to photon energy , then atom will absorb energy, otherwise, not. `E_(2) - E_(1) = 13.6 eV -(-54.4 eV)` `= 40.8 eV approx 41 eV` Similarly, `E_(3) - E_(1) = -6.04 eV - (-54.4 eV) = 48.36 eV` `E_(4) - E_(1) = -3.4 eV - (-54.4 eV) = 51 eV` `E_(3) - E_(2) = -6.04 eV - (-13.6 eV) = 7.56 eV` and so on, But note that `E_(2) - E_(1) ne 42 eV,E_(3) - E_(1) ne 42 eV, E_(4) - E_(1) ne 42 eV`and `E_(3) - E_(2) ne 42 eV`.For all possibilites, no difference in energy is an integer multiple of photon energy.Hence , photon A is not ABSORBED by this atom.But for Photon B, `E_(4) - E_(1) = 51 eV,` which means, Photon B can be absorbed by this atom. (e) Since total energy is equal to negative of kinetic energy in Bohr atom MODEL, we get `KE_(n) = - E_(n) = - (-(54.4)/(n^(2)) eV) = (54.4)/(n^(2)) eV` Potential energy is negative of twice the kinetic energy, which means, `U_(n) = -2KE_(n) = - 2((54.4)/(n^(2)) eV) = - (108.8)/(n^(2))eV` For a ground state, put n = 1, Kinetic energy is `KE_(1) = 54.4 eV` and Potential energy is `U_(1) = -108.8 eV` |
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