Optical deexcitation with energy higher than the excitation light?
Answers
Answered by
2
Hey mate!!
__________________________
⏩ Every line of the characteristic spectrum of an atom has not a perfectly defined energy (frequency), but a finite width (the higher the width, the shorter the lifetime of the excited state) .
⏩ Let's suppose we excite some type of atoms using monochromatic light with a very well defined frequency that is near the low-energy edge of the line width. If the atoms de-excitate later by spontaneous emission producing a line with the characteristic width, it means that sometimes the photon emitted in the de-excitation has more more energy than the photon used for the excitation.
Thanks for the question !
☺☺
__________________________
⏩ Every line of the characteristic spectrum of an atom has not a perfectly defined energy (frequency), but a finite width (the higher the width, the shorter the lifetime of the excited state) .
⏩ Let's suppose we excite some type of atoms using monochromatic light with a very well defined frequency that is near the low-energy edge of the line width. If the atoms de-excitate later by spontaneous emission producing a line with the characteristic width, it means that sometimes the photon emitted in the de-excitation has more more energy than the photon used for the excitation.
Thanks for the question !
☺☺
Monica695:
Hey aaisha44 please also thanks me two times because I have also thanks you 2 times please its a request
Answered by
0
HEY FRIEND
1. Every line of the characteristic spectrum of an atom has not a perfectly defined energy ( frequency ), but a finite with ( the higher the width, the shorter the lifetime of the excited state ) right?
2. Let's suppose we excite some type of atoms using monochromatic light with a very well defined frequency that is near the low-energy edge of the line width. If the atoms de-excitate later by spontaneous emission producing a line with the characteristic width, it means that sometimes the photon emitted in the de-excitation has more more energy than the photon used for the excitation. Where does this energy come from? Kinetic energy of the atoms? In that case, does it mean that we could reduce the temperature of the material this way?
HOPE THIS HELPS YOU.....
BY
aashiritika @ g.mail.com
1. Every line of the characteristic spectrum of an atom has not a perfectly defined energy ( frequency ), but a finite with ( the higher the width, the shorter the lifetime of the excited state ) right?
2. Let's suppose we excite some type of atoms using monochromatic light with a very well defined frequency that is near the low-energy edge of the line width. If the atoms de-excitate later by spontaneous emission producing a line with the characteristic width, it means that sometimes the photon emitted in the de-excitation has more more energy than the photon used for the excitation. Where does this energy come from? Kinetic energy of the atoms? In that case, does it mean that we could reduce the temperature of the material this way?
HOPE THIS HELPS YOU.....
BY
aashiritika @ g.mail.com
Similar questions
Computer Science,
7 months ago
Science,
1 year ago