Question

The wavelength of the first line of balmer series in hydrogen spectrum is 6562.8 Å.Calculare the ionisation potential of hydrogen and also the wave length of the first line of lyman series​

Answer 1

Solution:-

•The wavelength of the different lines of the Balmer series are given by

$\sf \: \: \: \dfrac{1}{ \lambda} = R_H \bigg( \dfrac{1}{ {2}^{2} } - \dfrac{1}{{n_i}^{2} } \bigg) \: \: where \: n_i = 3 \: ,4, \: 5.....$

⠀⠀⠀⠀⠀⠀⠀⠀

$\bull\sf Let \: \lambda_1 \: be \: the \: wavelength \: of \: first \: line \: of$

⠀⠀⠀⠀⠀⠀⠀⠀

Balmer series (ni=3).Then,

$\sf \: \: \dfrac{1}{\lambda_1} = R_H \bigg( \dfrac{1}{ {2}^{2} } - \dfrac{1}{ {3}^{2} } \bigg) = \dfrac{5R_H}{36}$

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: or \: \: R_H = \dfrac{36}{5\lambda_1}$

$\sf \: here \: \lambda_1 = 6562.8Å = 6562.8 \times {10 }^{ - 10} m$

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf R_H = \dfrac{36}{5 \times 6562.8 \times {10}^{ - 10}m }$

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: \: \: \: \: = 1.0971 \times {10}^{7} {m}^{ - 1}$

The energy of the electron in nth energy state of H-atom.

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: \: \: E_n = - \dfrac{R_H \: h_c}{ {n}^{2} }$

⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, ionization energy of H-atom

$\sf \: \: = E_ \infty - E_1$

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: = - \dfrac{R_H \: h_c}{ { \infty }^{2} } - \bigg( - \dfrac{R_H \: h_c}{ {1}^{2} } \bigg)$

$\sf \: \: = R_H \: h_c$

⠀⠀⠀⠀⠀⠀⠀⠀

$\: \: \sf = 1.0971 \times {10}^{7} \times 6.62 \times {10}^{ - 34} \times 3 \times {10}^{8}$

$\sf \: \: = 21.788 \times {10}^{ - 19} j$

$\sf \: \: = \dfrac{21.788 \times {10}^{ - 19} }{1.6 \times {10}^{ - 19} }$

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: = 13.6eV$

⠀⠀⠀⠀⠀⠀⠀⠀

The ionization potential is numerically equal to the ionization energy.

Therefore ,ionization potential of H-atom =13.6eV

⠀⠀⠀⠀⠀⠀⠀⠀

The wave length of the different lines of Lyman series are given by

$\sf \: \: \: \dfrac{1}{ \lambda} = R_H \bigg( \dfrac{1}{ {1}^{2} } - \dfrac{1}{{n_i}^{2} } \bigg) \: \:where \: n_i =2 \: ,3, \: 4.....$

$\bull\sf Let \: \lambda_1 \: be \: the \: wavelength \: of \: first \: line \: of$

Lyman series (ni=2).Then,

⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: \dfrac{1}{\lambda_1} = R_H \bigg( \dfrac{1}{ {1}^{2} } - \dfrac{1}{ {2}^{2} } \bigg) = \dfrac{3R_H}{4}$

$\sf \: \: \lambda_1 = \dfrac{4R_H}{3}$

$\sf \: \: \: \: = \dfrac{4}{3 \times 1.0971 \times {10}^{7} }$

⠀⠀⠀⠀⠀⠀⠀⠀

$\: \: \: \sf = 0.6875 \times {10}^{ - 10}$

⠀= 1215.3Å