Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A cation M n+ has a single electron. The highestenergy line in its emission spec

ID: 689537 • Letter: A

Question

A cation Mn+ has a single electron. The highestenergy line in its emission spectrum occurs at a frequency of8.225×1016 Hz. Identify the ion by its atomicsymbol (Answer 1) and its charge "n" (Answer 2).
E = -2.18*10-18 J (Z2/n2)=hv E = 6.626*10-34 J*s(8.225*1016 Hz) E = 5.45*10-17 J Z2= (En2)/ 2.18*10-18 J     = [(5.45*10-17 J) *(1)2]/ 2.18*10-18 J Z = 25 Z = 5 therefore, the element is B and the charge is+1 however when inputting my answer it says its wrong, pleasehelp! A cation Mn+ has a single electron. The highestenergy line in its emission spectrum occurs at a frequency of8.225×1016 Hz. Identify the ion by its atomicsymbol (Answer 1) and its charge "n" (Answer 2).
E = -2.18*10-18 J (Z2/n2)=hv E = 6.626*10-34 J*s(8.225*1016 Hz) E = 5.45*10-17 J Z2= (En2)/ 2.18*10-18 J     = [(5.45*10-17 J) *(1)2]/ 2.18*10-18 J Z = 25 Z = 5 therefore, the element is B and the charge is+1 however when inputting my answer it says its wrong, pleasehelp!     = [(5.45*10-17 J) *(1)2]/ 2.18*10-18 J Z = 25 Z = 5 therefore, the element is B and the charge is+1 however when inputting my answer it says its wrong, pleasehelp!

Explanation / Answer

We Know that :        Energy of electron forthe nth orbit of hydrogen like species is :           E = -2.18 * 10-18 J ( Z2 / n2 ) = hv                = -2.18 * 10-18 (  Z2 /n2 ) = 6.627 x 10^-34 J-s x 8.225 x1016 s-1                For the maximum frequency of the Atom the electronic ispresent at 1 st orbit i.e. n =1         Z2 = 6.627 x 10^-34 J-s x 8.225 x1016 s-1 x (1)2 / 2.18 * 10-18               = 25            Z = 5 The Atomic number of the Atomis 5 so it is Boron with symbol is B.           As ithas only one electron the charge on the B is +4 i.e. B+4                For the maximum frequency of the Atom the electronic ispresent at 1 st orbit i.e. n =1         Z2 = 6.627 x 10^-34 J-s x 8.225 x1016 s-1 x (1)2 / 2.18 * 10-18               = 25            Z = 5 The Atomic number of the Atomis 5 so it is Boron with symbol is B.           As ithas only one electron the charge on the B is +4 i.e. B+4
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote