Topics: Neutron, Atom, Proton Pages: 8 (2002 words) Published: January 20, 2013
TMA 07. Bernadett Kulcsar

Question 1.)

i) Eph=2.34 eV

ii) The energy transmitted between E7 and E5, when the electron jumps down from the E7, -2.52 eV state to the E5, -4.86 eV state.

b.) (i) The C5+ ion is a hydrogen-like ion, because it has lost 5 electrons and so it has one single electron in 1s orbital, just like the H atom. The C 5+ ion of the 12C isotope also has a nucleus containing 6 protons, and 6 neutrons each p+ contains 2 u and 1 d valence quarks, each neutron, n0, contains 2 d and 1 u quarks.The proton and neutrons also contain gluons holding the quarks together inside the proton, with the strong interaction.

(ii) En=( -13.60 eV x Z2) / n2, this is the equation to obtain the energy levels, where the ground state energy of hydrogen is -13.60 eV (4 sig figs), Z is the charge (equivalent to the atomic number ) and n is the principal quantum number, can be any positive whole number.So the energy levels are:

The atomic number of the C5+ ion is 6.

(n=1) E1=(-13,60 x 62)/1=-489.6 eV (4 sig figs)

(n=2) E2=(-13.60 x 62)/22=(-13.60 x 36)/4=-122.4 eV

(n=3) E3=(-13.60 x 36)/9=-54.40 eV

(n=4) E4 =(-13.60 x 36)/16=-30.60 eV

(n=5) E5=(-13.60 x 36)/25=-19.58 eV


Econtinuum = 0 eV

E5=-19.58 eV

EnergyE4=-30.60 eV

E3=-54.40 eV

E2=-122.40 eV

E1=-486.60 eV

where Econtinuum represents the state where n tends to infinity where the electron is free and not bound by the nucleus.


(i) The adsorption of a photon with the energy E5 - E3 = -19.58+54.40=34.82 eV can cause the ion’s electron energy to undergo a transition from the E3 and E5 levels.

(ii) The C5+ ion absorbs a photon of energy 500 eV.(equal to 0.5000 KeV) That energy is larger than any of the above calculated energy levels, it will become a 6+ positive ion because the photon energy is greater energy than the binding energy for all the energy levels, so that if the electron were in any energy level and absorbed the photon it would have enough energy to escape the attraction of the protons and wouldn’t be bounded anymore, so the ion would be further ionized by loosing the electron.

iii) The single electron in the C5+ ion experiences the full electric charge of the nucleus, but for either of the electrons of the C4+ ion with 2 electrons it would sometimes see the full charge of the nucleus, and then again sometimes less than that because the charge of the other electron would really be spread over the entire electron orbit and should cancel some of the nuclear charge. This is referred to as screening.

Question 2.)

a) Isotope A is 157N ,with atomic number of 7 and mass number of 15.

Isotope B is 23592U,with atomic number of 92 and mass number of 235.

Isotope C is 23692U,with atomic number of 92 and mass number of 236.

Both the charge and the mass are conserved,so they have to be equal on both sides.

i) Isotope A + 11p → 158O + 10n ; so mass(number of barions,which means protons and neutrons together) is A+1=15+1,so the mass number of isotope A has to be 15 too.Let’s see the charge,the number of protons in this case ,A + 1=8 + 0,so a has to be 7. I found Nitrogen with atomic number 7.


Isotope B +10n → isotope C→9942 Mo +13350Sn + 4 10n ; the mass number on the right side is 99+133+4=236,that is for isotope C. For isotope B it is one less, because there is one neutron as well, so the mass number of...
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