Question

Calculate the wavelength of a photon emitted by a hydrogen atom when its electron drops from n=5 state to n=3?

Answer 1

wavelength equals speed of light / frquency of oscillation....luckily this is the easiest way to do it if you know calculus, so we do not know frequency but we do know the change in state level which is amplitude. and there is a equation relating frequency and amplitude. frequency times wavelength equals amplitude. so if we rearrange we get frequency equals amplitude divided by wavelength...however we dont know wavelength but we do know the speed of light and plancks constant.. plancks constant is 6.62 x 10 ^-34 the equation to relate the planck and wavelength is wavelegth equals plancks constant divided by speed of light which is 3.00x 10 ^-8 . Use these equatuions to find wavelength Message me back right away so I know you're doing the problem well or if you need any more help... BTW SORRY FOR MISSPELLING TOO LAZY TO CORRECT THEM

Answer 2

this is what I did.
2.18 X 10^ 18 (1/5 ^2 - 1/3^2 )
= 3.03 X 10 ^ -19

THEN

(3.00 X 10 ^ 8) ( 6.63 X 10 ^ -34) / 3.03 X 10 - 19

= 656 nm

@_@

Answer 3

wrong. you have to consider mass as well if you're doing it that way. Can you think of an equation with mass in it? unless i'm reading what you did incorrectly

Answer 4

I was using Rydbergs constant then (3.00X10^8)(planck's constant)/(previous answer) = 656 nm

Answer 5

Check your physics book..somewhere in there there will be an equation with mass. if you can't get it in 7 minutes i'll tell you myself what it is. cool?

Answer 6

yeah

Answer 7

are you doing mastering physics or something? or just homework on a sheet of paper to turn in?

Answer 8

Im studyimg the end of chapter question but the answers arent in the back of the book lol

Answer 9

mRZ/λ = RZ2(m/n12 - m/n22)

where
m is the mass of the object
λ is the wavelength of the photon (wavenumber = 1/wavelength)
R = Rydberg's constant (1.0973731568539(55) x 107 m-1)
Z = atomic number of the atom
n1 and n2 are integers where n2 > n1.

Answer 10

quite similar but different

Answer 11

are you there?

Answer 12

@elvisG1993 I think there are a number of errors in your arguments, please explain if I'm incorrect ^_^

First, frequency times wavelength equals velocity, not amplitude.

Second, wavelength is not Planck's constant divide by c - for starters the units don't work out (you end up with kg m^2, not meters), and that would also imply that every wavelength is constant, since those two values are constants of nature.

Lastly, your Rydberg equation that you posted is not clear. If you simplify it down as written, there is no actual Rydberg constant in the equation because they both cancel out; the masses cancel out (I'm not entirely sure what the masses are for in your equation), and the dimensional analysis on the left gives units of 1/m and on the right is unitless. The wavenumber also is equal 2 pi / wavelength, not just the reciprocal of it, but I'm similarly unsure of its use here.


I believe the problem can be solved by applying a normal Rydberg formula without tinkering with it

\[\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\]
\[n_2=5
\\ n_1=3
\\ R=1.9074 \times 10^7 m^{-1}\]
This straightaway gives you 1 over the wavelength of the emitted photon.