Chemistry homework

Question

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=7n=7 to the level 𝑛=2.

Answer 1

Using the Equation first to obtain the delta E (energy lost in the transition of the electron from n=7 to n=2)

Delta E = Energy(final) - Energy(initial)

Energy Initial = the constant -2.178 x 10^-18J/7^2 where n=7 is the energy level where the electron started its transition.

Energy Final - -2.178 x 10^-18 J/2^2 where the electron finishes its transition at energy level n=2

Energy Initial = -4.445 x 10^-20 J

Energy Final = -5.445 x 10^-19 J

Delta Energy. = -5.445 x 10^-19 - (-4.445 x 10^-20) = -5.0 x 10^-19 J the negative number indicates energy was lost going from its excited state (n=7) to its ground state (n=2)

Eliminating the negative sign from the Delta energy as it only indicates whether energy was lost or gained in the electron transition, you can now use this equation to find v (frequency) E=hv where E is the energy you found in the electron transition, h = Planck's constant 6.63 X 10^-34 J.S and our unknown to find is the frequency (v)

5.0 x 10^-19 J = (6.63 x 10^-34 J.S)(v)

solving for v(frequency)

v = 5.0 x 10^-19J/6.63 x 10^-34J.S = 7.54 x 10^14 s^-1

Now using the relationship C(speed of light, 3.0 x 10^8m/s) = Lambda(wavelength) x v(frequency)

you know the speed of light and you know the v(frequency) just calculated above so:

3.0 x 10^8m/s = Lambda x 7.54 x 10^14 s^-1

solving forLambda(wavelength)

lambda = 3.0 x 10^8 m/s/7.54 x 10^14 s^-1 the units of S(seconds cancel you have lambda =

3.98 x 10^-7 m

now convert 3.98 x 10^-7m to nm = 3.98 x 10^2 nm = 398 nm