Prove that the ionisation energy of hydrogen atom is 13.6 eV.

Solution

We know that,

Energy,

E = - \frac{2 π^{2} {mK}^{2} Z^{2} e^{4}}{n^{2} h^{2}} (n = 1, Z = 1)

Energy required or the amount of work done to remove the electron from it's ground state to a higher level is,

W = k^{2} \frac{2 π^{2} {me}^{4}}{h^{2}} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

Ionisation energy is the energy required to remove an electron from ground state to infinity.

Here,

n_{1} = 1, n_{2} = \infty

∴

W = k^{2} \frac{2 π^{2} {me}^{4}}{h^{2}} (\frac{1}{1} - \frac{1}{\infty})

= k^{2} \frac{2 π^{2} {me}^{4}}{h^{2}}

\Rightarrow

W = \frac{(9 \times 10^{9})^{2} \times 2 (3.142)^{2} \times 9 \times 10^{- 31} \times (1.6 \times 10^{- 19})^{4}}{(6.63 \times 10^{- 34})^{2}} J

= 21.45 \times 10^{- 19} J = \frac{21.45 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV = 13.6 eV .

Thus, the ionisation of hydrogen atom is 13.6 eV.

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