Question

Calculate the wavelength of light emitted from a mol of photons as they transition from in the hydrogen atom.

Answer 1

The question is incomplete, here is the complete question:

How do you calculate the wavelength of the light emitted from a mole of photons as they transition from the n = 4 to the n = 1 principal energy level in the hydrogen atom. Recall that for hydrogen
`E_n=-2.18* 10^(-18)J((1)/(n^2))`

Answer: The wavelength of 1 mole of photons for the given transition is
`5.86* 10^(16)m`

Step-by-step explanation:

To calculate the change in energy, we use the equation:

`\Delta E=E_1-E_4`

Or,

`E_n=-2.18* 10^(-18)\left ((1)/(n_f^2)-(1)/(n_i^2)\right )`

where,

`n_i` = initial energy level = 4

`n_f` = final energy level = 1

Putting values in above equation, we get:

`\Delta E=-2.18* 10^(-18)\left ((1)/(1^2)-(1)/(4^2)\right )\\\\\Delta E=-2.044* 10^(-18)J`

To calculate the wavelength of light for 1 mole of photons, we use the equation:

`E=(N_Ahc)/(\lambda)`

where,

`N_A` = Avogadro's number =
`6.022* 10^(23)`

h = Planck's constant =
`6.625* 10^(-34)J.s`

c = speed of light =
`3* 10^8m/s`

`\lambda` = wavelength of light = ?

E = energy emitted =
`2.044* 10^(-18)J`

Putting values in above equation, we get:

`2.044* 10^(-18)J=(6.022* 10^(23)* 6.625* 10^(-34)J.s* 3* 10^8m/s)/(\lambda)\\\\\lambda=(6.022* 10^(23)* 6.625* 10^(-34)J.s* 3* 10^8m/s)/(2.044* 10^(-18)J)=5.86* 10^(16)m`

Hence, the wavelength of 1 mole of photons for the given transition is
`5.86* 10^(16)m`