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

Answer:

$\frac{1}{\lambda_{6-5}} \approx 752nm$

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

$\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )$

Therefore for initial drop of 5th to 2nd

$\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )$

Therefore for initial drop of 6th to 2nd

$\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )$

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

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

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

$\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}$

$\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}$

$\frac{1}{\lambda_{5-6}}=1.33*10^{-3}$

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

$\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}$

$\frac{1}{\lambda_{6-5}}=751.879nm$

$\frac{1}{\lambda_{6-5}} \approx 752nm$