To calculate the wavelength of the box, we can use the formula for the energy levels of a one-dimensional box:
\[ E_n = \frac{n^2h^2}{8mL^2} \]
where:
- \( E_n \) is the energy level
- \( n \) is the quantum number of the energy level
- \( h \) is the Planck's constant (6.626 x 10^-34 Js)
- \( m \) is the mass of the electron (9.11 x 10^-31 kg)
- \( L \) is the length of the box
Given that the electron requires a wavelength of 8080nm to excite it from n=2 to n=3 energy level, we can calculate the energy difference between these two levels:
\[ \Delta E = E_3 - E_2 = \frac{3^2h^2}{8mL^2} - \frac{2^2h^2}{8mL^2} \]
Substitute in the values of the constants and the given wavelength, \( \lambda = 8080nm = 8080 \times 10^{-9} m \), and solve for \( L \):
\[ \frac{hc}{\lambda} = \Delta E = \frac{9h^2}{8mL^2} - \frac{4h^2}{8mL^2} \]
\[ \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8080 \times 10^{-9}} = \frac{5h^2}{8mL^2} \]
\[ L = \sqrt{\frac{5h^2}{8m \times \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8080 \times 10^{-9}}}} \]
Calculating \( L \) gives:
\[ L = \sqrt{\frac{5(6.626 \times 10^{-34})^2}{8(9.11 \times 10^{-31}) \times \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8080 \times 10^{-9}}}} \]
\[ L \approx 5.97 \times 10^{-10} \, m \]
Therefore, the wavelength of the one dimensional box is approximately 597nm.
an electron in a one dimensional box required a wavelength of 8080nm to excite an electron from n=2 to n=3 energy level. calculate the wavelength of this box
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