91Ó°ÊÓ

The expression for the energy of photon is given by,

${\mathrm{E}}_{\mathrm{photon}}=\frac{\mathrm{hc}}{{\mathrm{λ}}_{\mathrm{light}}}$

Here ${\mathrm{E}}_{\mathrm{photon}}$is the energy of the photon,$\mathrm{c}$ is the speed of the light, is the Planck’s constant,

The value of $\mathrm{c}$in vacuum is $3.0×{10}^8\text{ }\mathrm{m}/\mathrm{s}$.

The expression for the energy of the oscillator is given by,

${\mathrm{E}}_{\mathrm{N}}={\mathrm{N³óÓ\neg }}_0+{\mathrm{E}}_0$...... (i)

Here $\mathrm{N}=0,1,2,\text{ }\mathrm{....}$

The angular frequency is given by,

${\mathrm{Ó\neg }}_0=\sqrt{\frac{{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$

Here ${\mathrm{Ó\neg }}_0$is the angular frequency, ${\mathrm{k}}_{\mathrm{s}}$is the spring constant, $\mathrm{m}$is the mass.

The ground state energy of the harmonic oscillator is given by,

${\mathrm{E}}_0=\frac{1}{2}{\mathrm{³óÓ\neg }}_0$ ....... (ii)

Substitute $\frac{1}{2}{\mathrm{³óÓ\neg }}_0$for ${\mathrm{E}}_0$into the equation (i)

$\begin{array}{c}{\mathrm{E}}_{\mathrm{N}}={\mathrm{N³óÓ\neg }}_0+\frac{1}{2}{\mathrm{³óÓ\neg }}_0\\ =\left(\mathrm{N}+\frac{1}{2}\right){\mathrm{³óÓ\neg }}_0\end{array}$

Now if we try to increase the energy by $\frac{1}{2}\mathrm{h}\sqrt{\frac{{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$$\left(\mathrm{or}\frac{1}{2}{\mathrm{³óÓ\neg }}_0\right)$then the new energy of the oscillator is,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{N}}={\mathrm{N³óÓ\neg }}_0+\frac{1}{2}{\mathrm{³óÓ\neg }}_0+\frac{1}{2}{\mathrm{³óÓ\neg }}_0\\ ={\mathrm{N³óÓ\neg }}_0+{\mathrm{³óÓ\neg }}_0\\ =(\mathrm{N}+1){\mathrm{³óÓ\neg }}_0\end{array}$

But the vibration energy levels of harmonic oscillator are quantized in the fashion of $\left(\mathrm{N}+\frac{1}{2}\right){\mathrm{³óÓ\neg }}_0$, So we can’t increase the energy by $\frac{1}{2}\mathrm{h}\sqrt{\frac{{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$$\left(\mathrm{or}\frac{1}{2}{\mathrm{³óÓ\neg }}_0\right)$ .