Question

Using the Bohr Model of the hydrogen atom, calculate the wavelength, frequency, and energy of the Humphreys gamma (n = 9  n = 6) spectral line. Would this spectral line be visible from the ground (you will have to investigate the transmission of the atmosphere)?

Answer 1

The wavelength is 2.27 μm.

The energy is
`-8.746*10^(-20)\ J`

The frequency is
`1.319*10^(14)\ Hz`

Step-by-step explanation:

Given that,

Number of spectral line n=9

Number of spectral line n=6

We need to calculate the wavelength

Using formula of wavelength

`(1)/(\lambda)=R_(h)((1)/(n_(f)^2)+(1)/(n_(i)^2))`

Put the value into the formula

`(1)/(\lambda)=1.097*10^(7)((1)/(36)+(1)/(81))`

`(1)/(\lambda)=440154.320988\ m`

`\lambda=0.0000022719\ m`

`\lambda=2.27*10^(-6)\ m`

`\lambda=2.27\ \mu m`

The wavelength is 2.27 μm.

We need to calculate the energy

Using formula of energy

`\Delta E=R_(h)((1)/(n_(f)^2)+(1)/(n_(i)^2))`

Put the value into the formula

`\Delta E=-2.18*10^(-18)*((1)/(36)+(1)/(81))`

`\Delta E=-8.746*10^(-20)\ J`

The energy is
`-8.746*10^(-20)\ J`

We need to calculate the frequency

Using formula of frequency

`f=(\Delta E)/(h)`

Put the value into the formula

`f=(8.746*10^(-20))/(6.62*10^(-34))`

`f=1.319*10^(14)\ Hz`

The frequency is
`1.319*10^(14)\ Hz`

Hence, The wavelength is 2.27 μm.

The energy is
`-8.746*10^(-20)\ J`

The frequency is
`1.319*10^(14)\ Hz`