The formula for the determination of energy is e=hv=hc/lamba where h is planck\'

Question

The formula for the determination of energy is e=hv=hc/lamba where h is planck's constant and c is the speed of light. what is the relationship between wavelength and energy?

(hydrogen) The rydberg equation has the form 1/lamba= Rh(1/nf2 - 1/ni2) where lamba is the wavelength in meters, Rh is the rydberg constant, nf is the final principal quantum (for the balmer series, which is in the visible spectrum, nf=2) and ni is the initial principal quantum number (n=3,4,5,6...). Calculate from data the wavelength in meters and 1/lamba in m1.

My data:

400 nm

433 nm

486 nm

656 nm

Of those above, which line corresponds to the transition n=3 to n=2, and from n=4 to n=2 and so on from n=6 to n=2.

Explanation / Answer

lamda = 400 nm = 400 x10^ -9 m = 4 x10^ -7 m

1/lmda = 1/4x10^-7 = 2.5 x10^ 6 m-1

433 nm = 4.33 x10^ -7 m ,

1/lamda = 1/4.33x10^-7 = 2.31 x10^ 6 m-1,

486 nm = 4.86 x10^ -7 m ,

1/lamda = 1/4.86 x10^ -7 = 2.057 x10^ 6 m-1,

656 nm = 6.56 x10^ 7 m ,

1/lamda = 1/6.56x10^ -7 = 1.525 x10^ 6 m-1,

for n =3 to n = 2 ,

wavenumber = 10967700(1/2^2 -1/3^2) = 1.524 x10^6 m-1

hence it corresponds to 656 nm

n= 4 to n= 2

wavenumber = 10967700(1/4-1/16) = 2.056 x10^6

it coresponds to 486 nm

for n= 6 to n= 2

wavenumber = 10967700(1/2^2 -1/6^2) = 2.44 x10^ 6

it corresponds to 400 nm

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