Question

When light of wavelength $248\text{nm}$ falls on a metal of threshold energy $3.0\text{eV}$, the de-Broglie wavelength of emitted electrons is $\stackrel{\circ }{A}$. (Round off to the nearest Integer).
$[\text{Use}:\sqrt{3}=1.73;h=6.63\times {10}^{-34}\text{J s};m_e=9.1\times {10}^{-31}\text{kg};c=3.0\times {10}^8{\text{m s}}^{-1};1\text{eV}=1.6\times {10}^{-19}\text{J}]$

Answer 1

Given,
$\text{Wavelength of light}\left(\lambda \right)=248\times {10}^{-9}m\text{Work function of metal}\left(w_0\right)=3\times 1.6\times {10}^{-19}\text{J}\text{By using the relation},E=w_0+K.E.\text{Where, E= total energy of electrons}\text{K.E.=kinetic energy of emitted electrons}\text{We know,}E=\frac{hc}{\lambda }\frac{hc}{\lambda }=w_0+K.E.\text{Substituting the values, we get}K.E=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{248\times {10}^{-9}}-3\times 1.6\times {10}^{-19}K.E.=3.2\times {10}^{-19}\text{J}\text{We know that,}P=\sqrt{2mK.E.}\text{Where, P=momentum of electron}\text{m= mass of electron}\text{Substituting the values, we get,}P=\sqrt{2\times 9.1\times {10}^{-31}\times 3.2\times {10}^{-19}}P=7.63\times {10}^{-25}\text{By using de-Broglie's relation,}\therefore \lambda =\frac{h}{P}=\frac{6.626\times {10}^{-34}}{7.63\times {10}^{-25}}\lambda =8.7\times {10}^{-10}\text{m}=8.7\stackrel{o}{A}\approx 9\stackrel{o}{A}$