Ncert Exercises & Solutions

2.14 How much energy is required to ionize an H atom if the electron occupies n = 5 orbits? Compare your answer with the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit).

The expression of energy is given by,

$E_{n} = \frac{- (2.18 x 10^{- 18}) Z^{2}}{n^{2}}$
Where,
Z = atomic number of the atom n
= principal quantum number
For ionization from

$n_{1}$ = 5 to

$n_{2} = \infty$ ,

$∆ E = E_{\infty} - E_{5}$

$= [\{\frac{- (2.18 x 10^{- 18} J) {(1)}^{2}}{{(\infty)}^{2}}\} - \{\frac{- (2.18 x 10^{- 18} J) {(1)}^{2}}{{(5)}^{2}}\}]$

$= (2.18 x 10^{- 18} J) (\frac{1}{(5^{2})}) (since \frac{1}{\infty} = 0)$

$= 0.0872 x 10^{- 18} J$

$∆ E = 8.72 x 10^{- 20} J$
Hence, the energy required for ionization from n = 5 to n =

$\infty$
Energy required for

$n_{1}$ = 1 to n =

$\infty$ , is 8.72 ×

$10^{- 20}$ J.

$∆ E = E_{\infty} - E_{1}$

$= [\{\frac{- (2.18 x 10^{- 18}) {(1)}^{2}}{{(\infty)}^{2}}\} - \{\frac{- (2.18 x 10^{- 18}) {(1)}^{2}}{{(1)}^{2}}\}]$

$= (2.18 x 10^{- 18} J) (1 - 0)$

$= 2.18 x 10^{- 18} J$
Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

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