Question

What is the wavelength, in nm, of a photon emitted during a transition from the n = 5 state to the n = 2 state in the hydrogen atom ?

Answer 1

This should be a transition in the so called "Balmer Series":

(Picture from Ohanian Physics)

You can use the fact that a photon emitted during the transition from n = 5 to n = 2 will carry an energy `E` equal to the difference between the energies of these two states.

Knowing this, you can relate the energy of the photon with the frequency `nu` through

`E=hnu`

(`h` is Planck's constant.)

For any state corresponding to `n` in the hydrogen atom, you get

`E_n= -"13.6 eV"/n^2`,

where `-"13.6 eV"` is the approximate ground-state energy of the hydrogen atom.

So:

`E_2 = -"13.6 eV"/4 = -"3.4 eV" = -5.44*10^-19` `"J"`

`E_5 = -"13.6 eV"/25 = "0.544 eV" = -8.7*10^-20` `"J"`

So,

`DeltaE = 4.57*10^-19` `"J"`.

In `E=hnu`,

`nu=(4.57*10^-19 "J")/(6.63*10^-34 "J" cdot"s") = 6.892*10^14 "s"^(-1)`,

but `c=lambdanu` (`c` is the speed of light);

So,

`lambda=(3*10^8 "m/s")/(6.892*10^14 "s"^(-1)) xx (10^9 "nm")/("1 m") = ul"435 nm"`

Answer 2

`lambda=433.936 7nm`

Explanation:

`lambda=(hc)/E_p`

`E_p` is the energy of photon.

Since, `E_p=(2pi^2m_eK^2Z^2e^4)/h^2(1/n_2^2-1/n_5^2)`

`K=1/(4piepsilon_0), n_2=2, n_5=5, Z=1`

`lambda=(800*h^3*c*epsilon_0^2)/(21*m_e*e^4)`

Putting in the values of physical constants

`lambda=433.9367 nm`

Here, `lambda=`wavelength of photon

`h=`Planck's constant

`m_e=`mass of electron

`c=`speed of light in vacuum

`e=`elementary charge

`Z=`Atomic No. [1 for hydrogen]