Respuesta :
Answer:
(a). The ground state energy is 16.56 eV.
(b). The wavelength of the emitted photon is 4.67 nm.
Explanation:
Given that,
Size l= 0.15 nm
We need to calculate the ground state energy
Using formula of energy state
[tex]E_{n}=\dfrac{n^2h^2}{8m_{e}L^2}[/tex]
For ground state,
[tex]E_{1}=\dfrac{(6.6\times10^{-34})^2}{8\times9.1\times10^{-31}\times(0.15\times10^{-9})^2}[/tex]
[tex]E_{1}=2.65\times10^{-18}\ J[/tex]
[tex]E_{1}=\dfrac{2.65\times10^{-18}}{1.6\times10^{-19}}[/tex]
[tex]E_{1}=16.56\ ev[/tex]
The ground state energy is 16.56 eV.
(b). We need to calculate the energy  difference between 5th excited and 3rd excited state
[tex]\Delta E=E_{5}-E_{3}[/tex]
[tex]\Delta E=(5^2-3^2)\dfrac{h^2}{8m_{e}l^2}[/tex]
[tex]\Delta E=16\times16.56[/tex]
[tex]\Delta E=264.96\ ev[/tex]
We need to calculate the wavelength of the emitted photon
Using formula of wave length
[tex]\Delta E=\dfrac{hc}{\lambda}[/tex]
[tex]\lambda=\dfrac{hc}{\Delta E}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{1240}{264.96}[/tex]
[tex]\lambda=4.67\ nm[/tex]
The wavelength of the emitted photon is 4.67 nm.
Hence, This is the required solution.
