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+4Related Questions
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