Question

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Ans.: 9.0110 111 diu J.LITU
Q5. LASER action occurs by stimulated emission from an excited state to a state of energy 30.5eV. If the
wavelength of LASER light emitted is 690 nm. What is the energy of the excited one? ​

Answer 1

Answer:

can't understand the question please write it again

Answer 2

Answer:

The required value of E2 = 8.71eV

Explanation:

We know that ∆E = 12400/λeV

and it is given that :

E1 = 30.5eV

λ = 690nm

Now we have to find out the energy present in the excited state , i.e, E2

therefore 690nm = 6900 A°

therefore, ∆E = 12400/6900 = 1.79ev

and ∆E = E1 - E2

or, 1.79 = 30.5 - E2

or, E2 = 28.71 eV

Therefore  the  required energy of the excited one is = 28.71eV

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