Achieve HW #8

[Briana Chavez 3F]

"Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n=3 to the level n=1."

Hello everyone! I understand how to do the first step of this problem, but I am confused about what I should be plugging in afterward. I used 1/lambda = R(1/(n1^2) - 1/(n2^2)), which resulted in (1.0974x10^7 m)(1/1^2 - 1/3^2) = 9.75x10^6 m. What am I supposed to do after this?

[Kathleen Wijaya 3L]

Since you were using the equation for 1/wavelength, you should take the reciprocal of 9.75x10^6 m to get wavelength. Then, convert to nanometers. Hope this helps!

[Claire Kim 1F]

"Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n=3 to the level n=1."

Hello everyone! I understand how to do the first step of this problem, but I am confused about what I should be plugging in afterward. I used 1/lambda = R(1/(n1^2) - 1/(n2^2)), which resulted in (1.0974x10^7 m)(1/1^2 - 1/3^2) = 9.75x10^6 m. What am I supposed to do after this?

You're on the right track. What I did is calculate when you plug in n=3 and n=1 (what you did), and then plug that value of v into the c=lambda(v). In this equation c is a constant and we now know v so therefore, we can easily solve for lambda.

[Aliza Hacking 1A]

Hi!! In addition to what everyone else is saying, you could also set up the equation so that you have lambda = c / (R(1/(n1^2) - 1/(n2^2)) ) from the start, which basically just combines that v = R(1/(n1^2) - 1/(n2^2)) equation with the c=lambda(v) equation. It's one less step and might help minimize confusion/human errors with the calculator along the way :))