Question

An electron in a hydrogen atom in its ground state absorbs energy equal to the ionization energy of $L{i}^{\text{+ 2}}$. The wavelength of the emitted electron is:

• A. $\text{3}\text{.32}\times {\text{10}}^{\text{- 10}}$ m
• B. $1.17\stackrel{\text{o}}{\text{A}}$
• C. $2.32\times {10}^{-9}\text{nm}$
• D. 3.33 pm

Answer 1

Answer: (B)

$1.17\stackrel{\text{o}}{\text{A}}$

Ionisation energy of lithium= $\frac{z^2}{n^2}\times 13.6$

$=\frac{3^2}{1^2}\times 13.6$

$=9\times 13.6$

$=122.4eV$

Ionisation energy of $H$ atom=$13.6eV$

Energy emitted= $122.4eV-13.6eV$

In Joules, energy emitted= $\frac{122.4eV-13.6J}{6.2419\times {10}^{18}}$

$=\frac{108.8J}{6.2419\times {10}^{18}}$

$=\frac{108.8}{6.2419}\times {10}^{-18}J$

$=\frac{1088}{62419}\times \frac{1000}{10}\times {10}^{-18}J$

$=\frac{1088}{62419}\times {10}^{-16}J$

$=1.743\times {10}^{-17}J$

Wavelength of photon using De Broglie Equation

$\lambda =\frac{hc}{E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.743\times {10}^{17}}$

$=\frac{6.62\times 3}{1.743}\times {10}^{-26+17}$

$=\frac{6.62\times 3}{1.743}\times {10}^{-9}$

$=1.17\times {10}^{-10}$

$=1.17\stackrel{o}{A}$