Question

4R* En = - The energy level of a Helium Ion with one electron is given by the expression: nº, when R* = 13.6eV What is the required energy to ionize such Helium ion from the ground state? What is the maximum wavelength of such ionization energy? How many spectral lines are in the emission spectrum of the Helium ions ionized by that radiation from the ground state?

Answer 1

There are 3 questions are in the given question

1. required energy to ionise Helium from ground state is -54.4eV solution i gave in the attachment image

2. maximum wavelength ($\lambda$) by using energy E=h$\nu$ formula

here $\nu$ = c/$\lambda$   here c= speed of light

wavelength is 0.354x10-24  solution part i did and attached that image

3. Number of spectral lines.

helium from its ground state it can excite maximum second level in that level 2s and 2p orbitals are present again in 2p due to degenerate orbital again splitting will takes place so from 2p 3 lines and from 2s one line total 4 spectral lines will from.

K S aus & Given data e [R = 13.6ev) Ena - n for Helium n=1 (for ground state) Ena - 4x13.6 su u =-54.4 from . rechired energy to ionise Heliumion ground state | En= -54.4 ev. - Manimum wave length as we know Echo Es hxc la [:20 =57 [ha planc constant] E = 6.627x10 X34100 +34 DILA 2= 6.627 x 14 x 3X100 19.881 x Topy 54.4 &= 0.3654x1024 = 3654x1028

in that » No. of spectral line of Heliumion. Helium Can manimum excited to and level level 25 & 2P orbitals are present in an 3 degenarate orbitals are present. - So Helium come from as 2 from ap < aby is is s one line one line one line and ale is one line So manimum it will form 4 line na Total spectral lines are = 4.

Thank you