Respuesta :
My Birthday is in April, So, electron of H atom is in n=4( pfund series).
Ionisation Energy= +13.6(z^2/n^2)
= +13.6(1/16)
= +0.85 eV
Ionisation Energy= +13.6(z^2/n^2)
= +13.6(1/16)
= +0.85 eV
Answer: Ionization energy = 145.9kJ/mol
Explanation: My birthday month is March, so taking level to be 3.
For calculating ionization energy we use Rydberg's expression for wavelength. After that we put the value for wavelength into Planck's Equation.
Rydberg's expression
[tex]\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)[/tex]
here
[tex]n_1^2=3\\n_2^2=\infty[/tex]
R = [tex]1.097\times10^7m^{-1}[/tex]
putting the values,
[tex]\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right)[/tex]
[tex]\frac{1}{\infty}=0[/tex]
therefore
[tex]\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{3^2}\right)[/tex]
[tex]\lambda= 8.204\times10^{-7}m[/tex]
Now, using planck's equation
[tex]E=h\nu[/tex] and
[tex]\nu=\frac{c}{\lambda}[/tex]
where h = [tex]6.626\times10^{-34}Js[/tex] (Planck's constant)
c = [tex]3\times10^8m/s[/tex] (Speed of light)
Putting value of [tex]\nu[/tex] in Energy formula, we get
[tex]E=\frac{hc}{\lambda}[/tex]
[tex]E=\frac{(6.626\times10^{-34}Js)(3\times10^8m/s)}{8.204\times10^-^7m}[/tex]
[tex]E=2.4229\times10^{-19}J/atom[/tex]
This amount of energy is required to remove 1 electron from 1 H-atom. Now to calculate the amount of energy required to remove 1 mole of electrons from 1 H-atom, we multiply it by Avagadro's Number
[tex]N_A=6.022\times10^{23}atoms/mol[/tex]
[tex]E=(2.4229\times10^{-19}J/atom)\times (6.022\times10^{23}atoms/mol)[/tex]
[tex]E=14.5907\times10^4J/mol[/tex]
[tex]E=145.907kJ/mol[/tex]
