Respuesta :
Explanation:
It is given that,
Length of one dimensional box, [tex]l=3\ nm=3\times 10^{-9}\ m[/tex]
Wavelength of electron, [tex]\lambda=1\ nm=10^{-9}\ m[/tex]
(a) We need to find the quantum number of this electron. The energy of electron in one dimensional box is given by :
[tex]E=\dfrac{n^2h^2}{8ml^2}[/tex]
Also, energy is given by, [tex]E=\dfrac{hc}{\lambda}[/tex]
[tex]\dfrac{hc}{\lambda}=\dfrac{n^2h^2}{8ml^2}[/tex]
[tex]n^2=\dfrac{8cml^2}{h\lambda}[/tex]
c is the speed of light
m is the mass of electron
h is the Planck's constant
[tex]n^2=\dfrac{8\times 3\times 10^8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2}{6.63\times 10^{-34}\times 10^{-9}}[/tex]
[tex]n=\sqrt{29647.05}[/tex]
n = 172.18
or
n = 172
(b) For ground state energy, n = 1
[tex]E=\dfrac{(6.64\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2}[/tex]
[tex]E=6.72\times 10^{-21}\ J[/tex]
(c) We need to find the wavelength of a photon that is emitted in a transition from the energy level in part a to the first excited state (n=2). So,
[tex]\dfrac{hc}{\lambda}=E_{172}-E_2[/tex]
[tex]\dfrac{hc}{\lambda}=\dfrac{(172)^2h^2}{8ma^2}-\dfrac{4h^2}{8ma^2}[/tex]
[tex]\dfrac{hc}{\lambda}=\dfrac{h^2}{8ml^2}(29647.05-4)[/tex]
[tex]\dfrac{hc}{\lambda}=\dfrac{h^2}{8ml^2}\times 29643.05[/tex]
[tex]\lambda=\dfrac{8ml^2c}{h\times 29643.05}[/tex]
[tex]\lambda=\dfrac{8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2\times 3\times 10^8}{6.63\times 10^{-34}\times 29643.05}[/tex]
[tex]\lambda=1.0001\times 10^{-9}\ m[/tex]
Hence, this is the required solution.
