IDNLearn.com: Where your questions meet expert advice and community support. Discover thorough and trustworthy answers from our community of knowledgeable professionals, tailored to meet your specific needs.
Sagot :
The question is incomplete. The complete question is :
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.
Solution :
In an one dimensional box, energy of a particle is given by :
[tex]$E=\frac{n^2h^2}{8ma^2}$[/tex]
Here, h = Planck's constant
n = level of energy
= 6
m = mass of particle
a = box length
For n = 6, the energy associated is :
[tex]$\Delta E = E_6 - E_1 $[/tex]
[tex]$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right) $[/tex]
[tex]$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$[/tex]
We know that,
[tex]$E = \frac{hc}{\lambda} $[/tex]
Here, λ = wavelength
h = Plank's constant
c = velocity of light
So the wavelength,
[tex]$= \frac{hc}{E}$[/tex]
[tex]$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$[/tex]
[tex]$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$[/tex]
[tex]$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$[/tex]
[tex]$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$[/tex]
[tex]$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$[/tex]
[tex]$= 0.108 \times 10^{-9}$[/tex] m
= 108 pm
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
