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)?

Respuesta :

Answer:

The wavelength is 2.27 μm.

The energy is [tex]-8.746\times10^{-20}\ J[/tex]

The frequency is [tex]1.319\times10^{14}\ Hz[/tex]

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

[tex]\dfrac{1}{\lambda}=R_{h}(\dfrac{1}{n_{f}^2}+\dfrac{1}{n_{i}^2})[/tex]

Put the value into the formula

[tex]\dfrac{1}{\lambda}=1.097\times10^{7}(\dfrac{1}{36}+\dfrac{1}{81})[/tex]

[tex]\dfrac{1}{\lambda}=440154.320988\ m[/tex]

[tex]\lambda=0.0000022719\ m[/tex]

[tex]\lambda=2.27\times10^{-6}\ m[/tex]

[tex]\lambda=2.27\ \mu m[/tex]

The wavelength is 2.27 μm.

We need to calculate the energy

Using formula of energy

[tex]\Delta E=R_{h}(\dfrac{1}{n_{f}^2}+\dfrac{1}{n_{i}^2})[/tex]

Put the value into the formula

[tex]\Delta E=-2.18\times10^{-18}\times(\dfrac{1}{36}+\dfrac{1}{81})[/tex]

[tex]\Delta E=-8.746\times10^{-20}\ J[/tex]

The energy is [tex]-8.746\times10^{-20}\ J[/tex]

We need to calculate the frequency

Using formula of frequency

[tex]f=\dfrac{\Delta E}{h}[/tex]

Put the value into the formula

[tex]f=\dfrac{8.746\times10^{-20}}{6.62\times10^{-34}}[/tex]

[tex]f=1.319\times10^{14}\ Hz[/tex]

The frequency is [tex]1.319\times10^{14}\ Hz[/tex]

Hence, The wavelength is 2.27 μm.

The energy is [tex]-8.746\times10^{-20}\ J[/tex]

The frequency is [tex]1.319\times10^{14}\ Hz[/tex]