Energy required to remove an electron from aluminum surface is 4.2 eV . If light of wavelength $2000 Å$…

Energy required to remove an electron from aluminum surface is 4.2 eV . If light of wavelength $2000 Å$ falls on the surface, the velocity of the fastest electron ejected from the surface will be
  1. $8.4 \times 10^5 \mathrm{~ms}^{-1}$
  2. $7.4 \times 10^5 \mathrm{~ms}^{-1}$
  3. $6.4 \times 10^5 \mathrm{~ms}^{-1}$
  4. $8.4 \times 10^6 \mathrm{~ms}^{-1}$

Solution

$\phi_{\mathrm{o}}=4.2 \mathrm{eV}, \lambda=2000 Å, \mathrm{v}_{\max }=?$ $\therefore \mathrm{E}=\frac{12400}{\lambda(\mathrm{in} Å)} \mathrm{eV}=\frac{12400}{2000}=6.2 \mathrm{eV}$ By Einstein's photoelectric equation, $\mathrm{k}_{\max }=\mathrm{E}-\phi_0=6.2-4.2=2 \mathrm{eV}$ $\Rightarrow \frac{1}{2} \operatorname{mv}_{\max }^2=2 \times 1.6 \times 10^{-19}$ $\therefore \quad \mathrm{v}_{\max }=8.4 \times 10^5 \mathrm{~m} / \mathrm{s}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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