Respuesta :
Answer:
The initial state is 6 and the final state is 2.
Explanation:
Given, wavelength of photon, [tex]\lambda =4.103\times 10^{-7} m[/tex]
and the value of Plank's constant, [tex]h=6.626\times 10^{-34} m^{2}kg/s[/tex]
and the velocity of light, [tex]c=3\times 10^{8} m/s[/tex]
And, [tex]1eV=1.6\times10^{-19} J[/tex]
According to the Bohr model the nth level of electron will be,
[tex]E_{n} =\frac{-13.6}{n^{2} } eV[/tex]
In the emission the energy is ejected out so,
[tex]-\Delta E=E_{ni}-E_{nf} \\-\Delta E=-13.6(\frac{1}{n_{i} ^{2} }-\frac{1}{n_{f} ^{2} } )eV[/tex]
And the change in energy also,
[tex]\Delta E=\frac{hc}{\lambda }[/tex]
Equate above two values.
[tex]\frac{hc}{\lambda }=-13.6(\frac{1}{n_{i} ^{2} }-\frac{1}{n_{f} ^{2} } )eV[/tex]
Therefore,
[tex]\frac{1}{n_{i} ^{2} } -\frac{1}{n_{f} ^{2} }=-\frac{hc}{\lambda (13.6 eV)}[/tex]
Put all the variables in the above equation
[tex]\frac{1}{n_{i} ^{2} } -\frac{1}{n_{f} ^{2} }=-\frac{6.626\times 10^{-34}(c=3\times 10^{8} )}{13.6(4.103\times 10^{-7})(1.6\times10^{-19} } \\\frac{1}{n_{i} ^{2} } -\frac{1}{n_{f} ^{2} }=0.22[/tex]
Now by putting the value of ni=6 and nf=2 to check the equality of equation.
[tex]\frac{1}{2^{2} } -\frac{1}{6^{2} }=0.22[/tex]
By looking at the above the value of ni=6 and nf=2 will satisfy the equation.
