## N initial and N final

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Kelsey Ash 1D
Posts: 50
Joined: Wed Sep 18, 2019 12:16 am

### N initial and N final

When trying to find energy level n initial or n final I know you use the equation delta E=(-hR/nf^2)-(-hR/ni^2) but when ever I try to isolate out the variable I am looking for I am off by 1 negative sign every time and I don't know what I am doing wrong. Could someone please show the work for this equation when solving for n initial?

sarahforman_Dis2I
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: N initial and N final

Kelsey Ash 1D wrote:When trying to find energy level n initial or n final I know you use the equation delta E=(-hR/nf^2)-(-hR/ni^2) but when ever I try to isolate out the variable I am looking for I am off by 1 negative sign every time and I don't know what I am doing wrong. Could someone please show the work for this equation when solving for n initial?

Something important to remember is that the negative sign is conceptual. If the value of the energy is negative, that means that the electron went from a higher energy level to a lower (eg 4--> 2), but if the sign is positive, the electron went from a lower to a higher energy level (eg 2--> 4). Check to make sure your final and initial values for n are correct.

To show this I can write out the equation below.

Transition from n=4 to n=2: (-hR/4)- (-hR/16) Since hR is a constant, hR/4 will be larger than hR/16. Adding hR/16 to -hR/4 will yield a negative number.

Transition from n=2 to n=4: (-hR/16)- (-hR/4) Since hR is a constant, hR/4 will be larger than hR/16. Adding hR/4 to -hR/16 will yield a positive number.