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

The wavelength of light is 2772 μm.

Step-by-step 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+(1)/(2))`

Put the value into the formula

`E_(5)=6.63*10^(-34)*1580*10^(-2)(5+(1)/(2))`

`E_(5)=5.76147*10^(-32)\ J`

For n = 2,

`E_(2)=6.63*10^(-34)*1580*10^(-2)(2+(1)/(2))`

`E_(2)=2.61885*10^(-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)*10^(-32)`

`E=3.14262*10^(-32)\ J`

We need to calculate the wavelength

Using formula of wavelength

`\lambda=(h)/(√(2mE))`

Put the value into the formula

`\lambda=\frac{6.63*10^(-34)}{\sqrt{2*9.1*10^(-31)*3.14262*10^(-32)}}`

`\lambda=0.00277224685446\ m`

`\lambda=2772*10^(-6)\ m`

`\lambda =2772\ \mu m`

Hence, The wavelength of light is 2772 μm.