Calculating Hydrogen Bohr Orbit Transition and Series using Wavelength 410.7 nm

[Benzoate]

Homework Statement

Light of wavelength 410.7 nm is observed in emission from a hydrogen source. a) what transition between hydrogen bohr orbits is responsible for this radiation? b) to what series does this transition belong to?

Homework Equations

1/lambda=Z^2 *R(1/n^2-m^2) ,

n being the final transition and m being the initial transition state. R is the Rydberg constant and the Rydberg constant for hydrogen is 1.096776*10^7 m^-1 and Z being the atomic number.

The Attempt at a Solution

Since the the wavelength is light , I know that its going to be in the visible spectrum of the E-M spectrum and the Balmer series is always part o the visible spectrum. the initial transition energy state in the balmer series is always m=2. Therefore since I know the initial transition state , R , Z and the wavelength , I thought I could find the n, the final transition state. here are my calculations below:

1/lambda)*1/(R)= (1/n^2-1/m^2) =>(1/(4.10e-7 m))*(1/(1.096776e7 m^-1))= (1/(2^2)-1/(m^2)) => .2224=(1/4-1/(m^2))=> -1/m^2 = -(.222-.25) => m^2= 1/.028 = 5.97 m = 6

 

[learningphysics]

Looks right. But the final state, n = 2... and you're finding m which is the initial state... so the initial state is m = 6.

 

[m2003]

a) The transition between hydrogen bohr orbits responsible for this radiation is from the initial state m=2 to the final state n=6.
b) This transition belongs to the Balmer series.