Answer:
[tex]\lambda=4.86*10^{-7}m[/tex]
Explanation:
Using the given equation, we calculate the energy associated with the excited state [tex]n_i=8[/tex] and [tex]n_f=4[/tex]
[tex]E_n=\frac{-2.18*10^-18J(Z^2)}{n^2}[/tex]
Helium has an atomic number (Z) equal to 2, for n=8:
[tex]E_8=\frac{-2.18*10^{-18}J(2^2)}{8^2}\\E_8=-1.36*10{-19}J[/tex]
For n=4:
[tex]E_4=\frac{-2.18*10^{-18}J(2^2)}{4^2}\\E_4=-5.45*10{-19}J[/tex]
When an electron jumps from an energy level with greater energy [tex] E_i [/tex] to one with lower energy [tex] E_f [/tex] the wavelength of the emitted photon is given by:
[tex]\lambda=\frac{hc}{E_i-E_f}[/tex]
h is the Planck constant and c the speed of light in vaccum. So, we have:
[tex]\lambda=\frac{hc}{E_8-E_4}\\\lambda=\frac{6.63*10^{-34}J \cdot s(3*10^8\frac{m}{s})}{-1.36*10{-19}J-(-5.45*10{-19}J)}\\\lambda=4.86*10^{-7}m[/tex]
The wavelength of the protons will be given as [tex]\lambda=4.86\times10^{-7}m[/tex]
The energy associated with the excited state will be calculated by the following equation
[tex]E_n=\dfrac{-2.18\times10^{-11}(Z^2)}{n^2}[/tex]
Since Helium has an atomic number (Z) = 2, for n=8:
[tex]E_8=\dfrac{-2.18\times10^{-18}(2^2)}{8^2}[/tex]
[tex]E_8=-1.36\times10^{-19}J[/tex]
Now For n=4:
[tex]E_4=\dfrac{-2.18\times10^{-18}(2^2)}{4^2}[/tex]
[tex]E_4=-5.45\times10^{-19}J[/tex]
When an electron jumps from an energy level with greater energy to one with lower energy the wavelength of the emitted photon is given by:
[tex]\lambda=\dfrac{hc}{E_i-E_f}[/tex]
Here
h is the Planck constant= [tex]6.63\times10^{-34[/tex]
c the speed of light in a vacuum =[tex]3\times10^8[/tex]
[tex]\lambda=\dfrac{6.63\times10^{-34}\times3\times10^8}{(-1.36\times10^{-19})-(-5.45\times10^{-19})}[/tex]
[tex]\lambda=4.86\times10^{-7}m[/tex]
Thus the wavelength of the protons will be given as [tex]\lambda=4.86\times10^{-7}m[/tex]
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