Question

An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 435 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.

Answer 1

`(1)/(\lambda_(6-5)) \approx 752nm`

Step-by-step explanation:

From the question we are told that

Light wavelength
`\lambda_l=655nm`

Light wavelength atom fall
`x_L=5th-2nd`

Photon wavelength
`\lambda_p=435nm`

Photon wavelength atom fall
`x_P=^th-2nd`

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

`(1)/(\lambda)=R((1)/( \lambda _f^2)-(1)/(\lambda _i^2) )`

Therefore for initial drop of 5th to 2nd

`(1)/(\lambda_(5-2))=R((1)/(2^2)-(1)/(5^2) )`

Therefore for initial drop of 6th to 2nd

`(1)/(\lambda_(6-2))=R((1)/(2^2)-(1)/(6^2) )`

Generally we subtract (5th to 2nd) from (6th to 2nd)

`(1)/(\lambda_(5-2))- (1)/(\lambda_(6-2))=R((1)/(2^2)-(1)/(5^2) )-\frac{1}R((1)/(2^2)-(1)/(6^2) )`

`(1)/(\lambda_(5-2))- (1)/(\lambda_(6-2))=R((1)/(5^2)-(1)/(6^2) )`

`(1)/(\lambda_(5-2))- (1)/(\lambda_(6-2))=(1)/(\lambda_(5-6))`

`(1)/(\lambda_(5-6))=(1)/(4350nm)-(1)/(655nm)`

`(1)/(\lambda_(5-6))=1.33*10^(-3)`

Therefore for 6th to 5th stage is mathematically given by

`(1)/(\lambda_(6-5))=(1.33*10^(-3))^(-1)`

`(1)/(\lambda_(6-5))=751.879nm`

`(1)/(\lambda_(6-5)) \approx 752nm`