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Step-by-Step Solution
Step 1: List the Known Parameters
• Wavelength of incident light, $ \lambda = 660 \times 10^{-9}\,\text{m} $
• Work function of the metal, $ \phi = 1.0\,\text{eV} $
• Planck’s constant, $ h = 6.626 \times 10^{-34}\,\text{J \cdot s} $
• Speed of light, $ c = 3.0 \times 10^{8}\,\text{m/s} $
• Mass of an electron, $ m_e = 9.1 \times 10^{-31}\,\text{kg} $
• $ 1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J} $
Step 2: Calculate the Photon Energy
The energy of a photon of wavelength $ \lambda $ is:
$$
E_\text{photon} = \frac{hc}{\lambda}.
$$
Substituting the values:
$$
E_\text{photon}
= \frac{(6.626 \times 10^{-34}\,\text{J \cdot s}) \times (3.0 \times 10^{8}\,\text{m/s})}
{660 \times 10^{-9}\,\text{m}}
\approx 3.01 \times 10^{-19}\,\text{J}.
$$
Step 3: Convert the Work Function to Joules
The work function in eV is turned into Joules:
$$
\phi
= 1.0\,\text{eV}
= 1.0 \times (1.6 \times 10^{-19})\,\text{J}
= 1.6 \times 10^{-19}\,\text{J}.
$$
Step 4: Determine the Kinetic Energy of the Ejected Electron
From Einstein’s photoelectric equation:
$$
K.E. = E_\text{photon} - \phi.
$$
Substituting the values:
$$
K.E.
= 3.01 \times 10^{-19}\,\text{J}
- 1.6 \times 10^{-19}\,\text{J}
= 1.41 \times 10^{-19}\,\text{J}.
$$
Step 5: Calculate the de Broglie Wavelength
The de Broglie wavelength of an electron with kinetic energy $K.E.$ is given by:
$$
\lambda
= \frac{h}{\sqrt{2\,m_e\,K.E.}}.
$$
Substituting the values:
$$
\lambda
= \frac{6.626 \times 10^{-34}\,\text{J \cdot s}}
{\sqrt{2 \times (9.1 \times 10^{-31}\,\text{kg}) \times (1.41 \times 10^{-19}\,\text{J})}}
\approx 1.3 \times 10^{-9}\,\text{m}.
$$
Final Answer
The de Broglie wavelength of the ejected electron is
$1.3 \times 10^{-9}\,\text{m}$.
