A gas absorbs a photon of 355 nm and emits at two wave lengths. If one

1.	A gas absorbs a photon of 355 nm and emits at two wave lengths. If one of the emissions is at 680 nm, at what place the other emission occur?
Answer» The wavelength of absorbed radiation is related to emitted radiation as below- \(\frac{1}{\lambda\,absorbed}\) = \(\frac{1}{\lambda_1}\) + \(\frac{1}{\lambda_2}\) \(\therefore\) \(\frac{1}{355}\) = \(\frac{1}{680}\) + \(\frac{1}{\lambda_2}\) \(\frac{1}{\lambda_2}\) = \(\frac{1}{355}\) - \(\frac{1}{680}\) = \(\frac{325}{241400}\) \(\lambda_2\) = \(\frac{241400}{325}\) = 742.78 ≈ 743nm

A gas absorbs a photon of 355 nm and emits at two wave lengths. If one of the emissions is at 680 nm, at what place the other emission occur?

Answer»

The wavelength of absorbed radiation is related to emitted radiation as below-

\(\frac{1}{\lambda\,absorbed}\) = \(\frac{1}{\lambda_1}\) + \(\frac{1}{\lambda_2}\)

\(\therefore\) \(\frac{1}{355}\) = \(\frac{1}{680}\) + \(\frac{1}{\lambda_2}\)

\(\frac{1}{\lambda_2}\) = \(\frac{1}{355}\) - \(\frac{1}{680}\)

= \(\frac{325}{241400}\)

\(\lambda_2\) = \(\frac{241400}{325}\)

= 742.78 ≈ 743nm