Question

A single electron orbits a stationary nucleus of charge $+Ze$, where $Z$ is a constant and $e$ is the magnitude of electronic charge. It requires $47.2eV$ to excite the electron from the second Bohr orbit to third Bohr orbit. Find
(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 electromagnetic required to remove the electron in 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.

Answer 1

$\text{We Know,}$

$E=13.6Z^2(\frac{1}{n^2}-\frac{1}{m^2})$, E=Ionisation Energy

here $n=2,m=3$

$47.2=13.6Z^2(\frac{1}{2^2}-\frac{1}{3^2})$

$47.2=13.6Z^2(\frac{1}{4}-\frac{1}{9})$

$47.2=13.6Z^2\left(\frac{5}{36}\right)$

$Z^2=47.2\times 36\times \frac{1}{13.6\times 5}=25$

$Z=5$

$\text{Now,}$

$E_{3,4}=13.6\times 5^2(\frac{1}{3^2}-\frac{1}{4^2})$

$E_{3,4}=13.6\times 5^2(\frac{1}{9}-\frac{1}{16})$

$E_{3,4}=13.6\times 5^2\left(\frac{7}{144}\right)$

$E_{3,4}=13.6\times 25\times \frac{7}{144}=340\times \frac{7}{144}=\frac{2380}{144}$

$E_{3,4}=16.52eV$

$\text{Now,}$

$K{E}_1=13.6\times \frac{5^2}{1^2}=13.6\times 25=340eV$

$P{E}_1=-2K{E}_1=-680eV$

$\text{Angular momentum}=\frac{nh}{2\pi }=\frac{h}{2\pi }$

Now,

$r_1=0.529\times \frac{n^2}{Z}=0.529\times \frac{1}{5}=0.1058A^{\odot }$

$=10.5pm=1.058\times {10}^{-11}m$