Question

It is common for atomic physicists to quote a frequency in cm^−1, since f=c/λ=3× 10^10 cm/s/λ. The vibrational frequency for the interatomic distance in O2 is f=1580 cm^−1. What wavelength of light in micrometers is emitted when an O2 molecule jumps from the vibrational level (for which En=hf(n+1/2)) with n=5 to n=2?

Answer 1

Answer:

The wavelength of light is 2772 μm.

Explanation:

Given that,

Frequency = 1580 cm⁻¹

Higher state =5

Lower state = 2

We need to calculate the energy for n = 5

Using formula of energy

$E_{5}=hf(n+\dfrac{1}{2})$

Put the value into the formula

$E_{5}=6.63\times10^{-34}\times1580\times10^{-2}(5+\dfrac{1}{2})$

$E_{5}=5.76147\times10^{-32}\ J$

For n = 2,

$E_{2}=6.63\times10^{-34}\times1580\times10^{-2}(2+\dfrac{1}{2})$

$E_{2}=2.61885\times10^{-32}\ J$

We need to calculate the energy

Using formula of energy

$E=E_{5}-E_{2}$

Put the value into the formula

$E=(5.79623-2.63465)\times10^{-32}$

$E=3.14262\times10^{-32}\ J$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3.14262\times10^{-32}}}$

$\lambda=0.00277224685446\ m$

$\lambda=2772\times10^{-6}\ m$

$\lambda =2772\ \mu m$

Hence, The wavelength of light is 2772 μm.