Question

A single electron orbits around a stationary nucleus of charge +Ze. Where Z is a constant and e is the magnitude of the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit.
(a) the value of Z.
(b) the energy required to excite the electron from the third to the fourth Bohr orbit.
(c) the wavelength of the electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
(d) the kinetic energy, potential energy and the angular momentum of the electron in the first Bohr orbit.
(e) the radius of the first Bohr orbit.
(The ionization energy of hydrogen atom =13.6 eV, Bohr radius $=5.3\times10^{-11}m$, velocity of light $=3\times10^8 m/s.$ Planck's constant $=6.6\times10^{-34}J-s$)

Answer 1

(a) Given $E_3-E_2=47.2 eV$ Since $E_n \propto \frac{Z^2}{n^2}$ (for hydrogen like atoms) or $(-13.6)\bigg(\frac{Z^2}{9}\bigg)-\bigg[(-13.6)\bigg(\frac{Z^2}{4}\bigg)\bigg]=47.2$ Solving this equation, we get $Z=5$ (b) Energy required to excite the electron from 3rd to 4th orbit: $E_{3-4}=E_4-E_3$ $(-13.6)\bigg(\frac{25}{169}\bigg)-\bigg[(-13.6)\bigg(\frac{25}{9}\bigg)\bigg]=16.53eV$ (c) Energy required to remove the electron from first orbit to infinity (or the ionisation energy) will be : $E=(13.6)(5)^2 = 340 eV$ The corresponding wavelength would be. $\lambda=\frac{hc}{E}=\frac{6.6\times10^{-34}\times3\times10^8}{340\times1.6\times10^{-19}}$ $=0.0364\times10^{-7}m = 36.4\mathring{A} (d) In first orbit, total energy = - 340 eV kinetic energy = + 340 eV angul1 ar momentum $=\frac{h}{2\pi} $=\frac{6.6\times10^{-34}}{2\pi}$ $=1.05\times10^{-34}kg-m^2/s$ (e) $r_n \propto \frac{n^2}{Z}$ Radius of first Bohr orbit $r_1=\frac{r^H_1}{Z}=\frac{5.3\times10^{-11}}{5}$ $=1.06\times10^{-11}m$