Question

The energy levels of one‑electron ions are given by the equation En=(−2.18×10−18 J)(Z^2/n^2) where Z is atomic number and n is the energy level. The series in the He+ spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf=4 state is called the Pickering series, an important series in solar astronomy. Calculate the Pickering series wavelength associated with the excited state ni=8.

Answer 1

Answer:

$\lambda=4.86*10^{-7}m$

Explanation:

Using the given equation, we calculate the energy associated with the excited state $n_i=8$ and $n_f=4$

$E_n=\frac{-2.18*10^-18J(Z^2)}{n^2}$

Helium has an atomic number (Z) equal to 2, for n=8:

$E_8=\frac{-2.18*10^{-18}J(2^2)}{8^2}\\E_8=-1.36*10{-19}J$

For n=4:

$E_4=\frac{-2.18*10^{-18}J(2^2)}{4^2}\\E_4=-5.45*10{-19}J$

When an electron jumps from an energy level with greater energy $E_i$ to one with lower energy $E_f$ the wavelength of the emitted photon is given by:

$\lambda=\frac{hc}{E_i-E_f}$

h is the Planck constant and c the speed of light in vaccum. So, we have:

$\lambda=\frac{hc}{E_8-E_4}\\\lambda=\frac{6.63*10^{-34}J \cdot s(3*10^8\frac{m}{s})}{-1.36*10{-19}J-(-5.45*10{-19}J)}\\\lambda=4.86*10^{-7}m$