Light of photon energy 3.5 eV falls on a metal surface of work function 2.25 eV. The minimum de Broglie wavelength of the emitted photoelectrons is (Take \( m_e=9\times10^{-31}\,\mathrm{kg} \), \( h=6.6\times10^{-34}\,\mathrm{J\,s} \) and \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.1 nm
The fastest photoelectrons carry the maximum kinetic energy given by Einstein’s photoelectric equation, \( K_{\max}=h\nu-\phi_0 \). The de Broglie wavelength \( \lambda=h/p \) is shortest for the largest momentum, so these fastest electrons have the minimum wavelength. An electron of mass \( m \) and kinetic energy \( K \) has momentum \( p=\sqrt{2mK} \).
The maximum kinetic energy of the photoelectrons is:
\( \begin{aligned} K_{\max} &= h\nu-\phi_0 \\ &= 3.5\,\mathrm{eV}-2.25\,\mathrm{eV} \\ &= 1.25\,\mathrm{eV} \\ &= 1.25\times1.6\times10^{-19}\,\mathrm{J} \\ &= 2.0\times10^{-19}\,\mathrm{J} \end{aligned} \)
The momentum of these electrons is:
\( \begin{aligned} p &= \sqrt{2mK_{\max}} \\ &= \sqrt{2\times9\times10^{-31}\,\mathrm{kg}\times2.0\times10^{-19}\,\mathrm{J}} \\ &= \sqrt{3.6\times10^{-49}}\,\mathrm{kg\,m\,s^{-1}} \\ &= 6.0\times10^{-25}\,\mathrm{kg\,m\,s^{-1}} \end{aligned} \)
Their de Broglie wavelength is then:
\( \begin{aligned} \lambda_{\min} &= \frac{h}{p} \\ &= \frac{6.6\times10^{-34}\,\mathrm{J\,s}}{6.0\times10^{-25}\,\mathrm{kg\,m\,s^{-1}}} \\ &= 1.1\times10^{-9}\,\mathrm{m} \\ &= 1.1\,\mathrm{nm} \end{aligned} \)