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:

\[ \Delta E = \frac{h^2}{8mL^2} (n_f^2 - n_i^2) \]

Where:
- \( \Delta E \) 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:

\[ \Delta E = \frac{hc}{\lambda} \]

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 \( L \):

\[ \frac{h^2}{8mL^2} (3^2 - 2^2) = \frac{hc}{\lambda} \]

\[ \frac{h^2}{8mL^2} (9 - 4) = \frac{hc}{\lambda} \]

\[ \frac{h^2}{8mL^2} = \frac{hc}{\lambda} \]

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

Now we can plug in the values:

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

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

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

\[ L = 2.45 x 10^-34 m \]

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