91Ó°ÊÓ

Spring constant (k) = 2.0N/m

Particle: electron

Angular frequency of electron in this case is: $\sqrt{\frac{k}{m}}$

$$\begin{gathered} =\sqrt{\frac{2.0N/m}{9.11×{10}^{-31}kg}} \\ =\sqrt{2.1953×{10}^{30}} \\ =1.481×{10}^{15}rad/sec \end{gathered}$$

Now we know that energy level of quantum harmonic oscillator is $E=(n-\frac{1}{2})\frac{h}{2\mathrm{Ï€}}\mathrm{Ó\neg }$

For, first energy level, n = 1

$$\begin{gathered} E_1=\frac{h}{4\mathrm{Ï€}}\mathrm{Ó\neg } \\ =\frac{6.626×{10}^{-34}JH{z}^{-1}}{4×3.14}×1.481×{10}^{15}rad/sec \\ 7.809×{10}^{-20}J \end{gathered}$$

For second energy level, n = 2

$$\begin{gathered} E_2=\frac{3h}{4\mathrm{Ï€}}\mathrm{Ó\neg } \\ =\frac{3×6.626×{10}^{-34}JH{z}^{-1}}{4×3.14}×1.481×{10}^{15}rad/sec \\ 2.342×{10}^{-19}J \end{gathered}$$

For third energy level, n = 3

$$\begin{gathered} E_3=\frac{5h}{4\mathrm{Ï€}}\mathrm{Ó\neg } \\ =\frac{5×6.626×{10}^{-34}JH{z}^{-1}}{4×3.14}×1.481×{10}^{15}rad/sec \\ 3.9045×{10}^{-19}J \end{gathered}$$

If electron makes a jump from E₃to E_1.The energy of photon will be equal to the energy difference of these levels.

$⇒\frac{hc}{\mathrm{λ}}=E_3-E_1$

$$\begin{gathered} ⇒\mathrm{λ}=\frac{hc}{E_3-E_1} \\ SubstitutingValues \\ \mathrm{λ}=\frac{6.626×{10}^{-34}J/Hz×3×{10}^{8}m/s}{(3.9045×{10}^{-19})-(0.7809×{10}^{-19})J} \\ ⇒\mathrm{λ}=6.3638×{10}^{-7}m \\ ⇒\mathrm{λ}=636.38nm \end{gathered}$$