Question

The series limit for Balmer series of $H$ spectrum occurs at $3664\mathring{A}$. Calculate (a) ionization energy of $H$-atom.
(b) the wavelength of the photon that would remove the electron in the ground state of the $H$-atom.

• A. (a)$E=5.42\times {10}^{-19}J$
  (b)$\lambda =916\mathring{A}$
• B. (a)$E=1.42\times {10}^{-19}J$
  (b)$\lambda =16\mathring{A}$
• C. (a)$E=5.42\times {10}^{-19}J$
  (b)$wavenumber=27\times {10}^5{m}^{-1}$
• D. (a)$E=3.38eV$
  (b)$wavenumber=109\times {10}^5{m}^{-1}$

Answer 1

Answer: (A), (D)

(a)$E=5.42\times {10}^{-19}J$
(b)$\lambda =916\mathring{A}$
(a)$E=3.38eV$

(b)$wavenumber=109\times {10}^5{m}^{-1}$
Given series is Blamer series,i.e.,
$\lambda =3664A^o$ and thus, $n_1=2;n_2=\infty$
(a)$\because E$ photon of series limit $=\Delta E=E_{infty}-E_2=-E_2$
$=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{3664\times {10}^{-10}}$
or $E_2=-5.42\times {10}^{-19}J$
(b)$\because E_1=E_2\times n^2=-4\times 5.42\times {10}^{-19}J$
$=-21.68\times {10}^{-19}J$
Now for removal of electron from $I$ orbit $\frac{hc}{\lambda }=21.68\times {10}^{-19}$;
$21.68\times {10}^{-19}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{\lambda }$
$\therefore \lambda =916\mathring{A}$