Question

Ultraviolet light of wavelengths $800\stackrel{o}{A}$ and $700\stackrel{o}{A}$ when allowed to fall on hydrogen atoms in their ground states is found to liberate electrons with kinetic energies 1.8 eV and 4.0 eV, respectively. If the value of Planck's constant is $6.5\times {10}^{-X}$ [1 electrons volt $\left(eV\right)=1.6\times {10}^{-19}J]$. Find X?

Answer 1

The energy of incident photon
$=\left(hc/\lambda \right)$
If $W_i$ is the ionization energy and $E_k$ the kinetic energy of emitted electron, then we have
$\frac{hc}{{\lambda }_1}=W_1+E_{K_1}$
Therefore, for incident photon of wavelength
${\lambda }_1=800\stackrel{o}{A}=8\times {10}^{-8}m$
$\frac{hc}{\lambda }=W_1+E_{K_1}$ (i)
And for incident photon of wavelength
${\lambda }_2=700\stackrel{o}{A}=7\times {10}^{-8}m$,
$\frac{hc}{\lambda }=W_1+E_{K_2}$ (ii)
Subtracting Eqs. (ii) from (i), we get
$hc(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1})=E_{K_2}-E_{K_1}$
or $h=\frac{(E_{K_2}-E_{K_1}){\lambda }_1{\lambda }_2}{c({\lambda }_1-{\lambda }_2)}$
Here, $E_{K_1}=1.eV=1.8\times 1.6\times {10}^{-19}J$
and $E_{K_2}=4.0eV=4.0\times 1.6\times {10}^{-19}J$
Substituting given values, we get
$h=\frac{(4.8-1.8)\times 1.6\times {10}^{19}\times 8\times {10}^{-8}\times 7\times {10}^{-8}}{3\times {10}^{-8}(8\times {10}^{-8}-7\times {10}^{-8})}$
$=6.57\times {10}^{-34}Js$