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A gas absorbs a photon of 355 nm and emit at two wavelengths. If one of the emission is at 680

Answer» <html><body><p>325 nm<br/>743 nm<br/>518 nm<br/>1035 nm</p>Solution :As energies are additive, `E = E_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) + E_(2)` <br/> or `(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a>) = (hc)/(lamda_(1)) + (hc)/(lamda_(2)) or (1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2))` <br/> But `lamda = 355 nm, lamda_(1) = <a href="https://interviewquestions.tuteehub.com/tag/680nm-1911337" style="font-weight:bold;" target="_blank" title="Click to know more about 680NM">680NM</a>,` <br/> `:. (1)/(355) = (1)/(680) + (1)/(lamda_(2))` <br/> or `(1)/(lamda_(2)) = (1)/(355) - (1)/(680) = (325)/(355 xx 680)` <br/> or `lamda_(2) = (355 xx 680)/(325) = 745nm`</body></html>


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