An electron in a one dimensional box requires a wavelength of 80nm to excite an electron from n= 2 to the n=3 energy level. Calculate the wavof this box step by step?

Answer 1

To calculate the length of the box, we first need to determine the energy difference between the n=2 and n=3 energy levels. The energy difference between energy levels in a one-dimensional box is given by the formula:

$$\mathrm{\Delta }E=\frac{h^2}{8m{L}^2}(n_f^2-n_i^2)$$

Where:
- $\mathrm{\Delta }E$ is the energy difference between the two energy levels
- $h$ is the Planck constant (6.626 x 10^-34 J*s)
- $m$ is the mass of the electron (9.11 x 10^-31 kg)
- $L$ is the length of the box
- $n_f$ is the final energy level (3)
- $n_i$ is the initial energy level (2)

Given that the wavelength of the excitation is 80nm, we can calculate the energy difference using the formula:

$$\mathrm{\Delta }E=\frac{hc}{\lambda }$$

Where:
- $c$ is the speed of light (3 x 10^8 m/s)
- $\lambda$ is the wavelength (80 x 10^-9 m)

Now we can set these two equations equal to each other and solve for $L$:

$$\frac{h^2}{8m{L}^2}(3^2-2^2)=\frac{hc}{\lambda }$$

$$\frac{h^2}{8m{L}^2}(9-4)=\frac{hc}{\lambda }$$

$$\frac{h^2}{8m{L}^2}=\frac{hc}{\lambda }$$

$$L=\sqrt{\frac{h^2}{8m}\frac{\lambda }{c}}$$

Now we can plug in the values:

$$L=\sqrt{\frac{(6.626x{10}^{-}34{)}^2}{8(9.11x{10}^{-}31)}\frac{80x{10}^{-}9}{3x{10}^8}}$$

$$L=\sqrt{\frac{(4.395x{10}^{-}67)}{72.88}\frac{80x{10}^{-}9}{3x{10}^8}}$$

$$L=\sqrt{6.033x{10}^{-}69}$$

$$L=2.45x{10}^{-}34m$$

Therefore, the length of the box is approximately 2.45 x 10^-34 meters.