Question

Assume a hydrogen atom has an electron in the n-level corresponding to the month of your birthday (jan = 1, feb = 2, etc.) calculate the ionization energy of that hydrogen atom. hint: the ionization energy is defined as the energy needed to completely remove an electron from an atom, ion, or molecule.

Answer 1

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

Answer 2

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

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

here

$n_1^2=3\\n_2^2=\infty$

R = $1.097\times10^7m^{-1}$

putting the values,

$\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right)$

$\frac{1}{\infty}=0$

therefore

$\frac{1}{\lambda}=1.097\times10^7\left(\frac{1}{3^2}\right)$

$\lambda= 8.204\times10^{-7}m$

Now, using planck's equation

$E=h\nu$ and

$\nu=\frac{c}{\lambda}$

where h = $6.626\times10^{-34}Js$ (Planck's constant)

c = $3\times10^8m/s$ (Speed of light)

Putting value of $\nu$ in Energy formula, we get

$E=\frac{hc}{\lambda}$

$E=\frac{(6.626\times10^{-34}Js)(3\times10^8m/s)}{8.204\times10^-^7m}$

$E=2.4229\times10^{-19}J/atom$

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

$N_A=6.022\times10^{23}atoms/mol$

$E=(2.4229\times10^{-19}J/atom)\times (6.022\times10^{23}atoms/mol)$

$E=14.5907\times10^4J/mol$

$E=145.907kJ/mol$