Respuesta :
Answer:
(a) 1.2703×10⁻⁷ m
(b) 4.7636×10⁻⁸ m
(c) 2.5406×10⁻⁸ m
Explanation:
Given:
Width of the infinite well, L = 340 pm = 340×10⁻¹² m.
The formula for energy of the electron in nth state is:
[tex]E_n=\frac {n^2\times h^2}{8mL^2}[/tex]
The expression for the difference in energy between the levels having quantum numbers n(initial) to n(final) is:
[tex]\Delta E_n=\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}[/tex]
According to Planks theory:
E = hv
where, v is the frequency
Also,
Frequency×Wavelength = Speed of light
So,
[tex]E=\frac {hc}{\lambda}[/tex]
[tex]\lambda=\frac {hc}{E}[/tex]
Also, using energy from above formula as:
[tex]\lambda=\frac {hc}{\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}}[/tex]
[tex]\lambda=\frac {c\times {8mL^2}} {({n_f}^2-{n_i}^2)\times h}}[/tex]
For longest wavelength ni = 1 and nf = 2
m= mass of the electron = 9.1 ×10⁻³¹kg
c = 3×10⁸m/s
h = 6.625×10⁻³⁴J.sec
[tex]\lambda_{Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({2}^2-{1}^2)\times 6.625\times 10^{-34}}}[/tex]
Longest wavelength = 1.2703×10⁻⁷ m
For second longest wavelength ni = 1 and nf = 3
[tex]\lambda_{Second\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({3}^2-{1}^2)\times 6.625\times 10^{-34}}}[/tex]
Second longest wavelength = 4.7636×10⁻⁸ m
For third longest wavelength ni = 1 and nf = 4
[tex]\lambda_{Third\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({4}^2-{1}^2)\times 6.625\times 10^{-34}}}[/tex]
Third longest wavelength = 2.5406×10⁻⁸ m
