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Step-by-Step Solution
Step 1: List the Given Data
• Wavelength of incident light, \lambda = 248 \times 10^{-9}\,\text{m}
• Threshold energy of metal, W_0 = 3.0\,\text{eV}
• Planck’s constant, h = 6.63 \times 10^{-34}\,\text{J·s}
• Mass of electron, m_e = 9.1 \times 10^{-31}\,\text{kg}
• Speed of light, c = 3.0 \times 10^{8}\,\text{m/s}
• 1 eV = 1.6 \times 10^{-19}\,\text{J}
Step 2: Convert the Threshold Energy to Joules
Since 1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}, we have
W_0 = 3.0\,\text{eV} = 3.0 \times 1.6 \times 10^{-19}\,\text{J} = 4.8 \times 10^{-19}\,\text{J}.
Step 3: Calculate the Energy of the Incident Photon
Photon energy is given by
E_{\text{photon}} = \frac{hc}{\lambda}.
Substituting the values:
E_{\text{photon}}
= \frac{(6.63 \times 10^{-34}\,\text{J·s}) \times (3.0 \times 10^{8}\,\text{m/s})}{248 \times 10^{-9}\,\text{m}}.
Simplifying will give energy in Joules. Let’s perform the approximate calculation:
Numerator: 6.63 \times 10^{-34} \times 3.0 \times 10^{8} = 1.989 \times 10^{-25}\,\text{J·m}.
Denominator: 248 \times 10^{-9} = 2.48 \times 10^{-7}\,\text{m}.
Thus,
E_{\text{photon}} \approx \frac{1.989 \times 10^{-25}}{2.48 \times 10^{-7}}\,\text{J}.
E_{\text{photon}} \approx 8.0 \times 10^{-19}\,\text{J} \;(\text{approx.}).
Step 4: Calculate the Kinetic Energy of the Emitted Electron
The emitted electron’s kinetic energy ( K.E. ) is the photon energy minus the threshold energy:
K.E. = E_{\text{photon}} - W_0.
Substituting approximate values:
K.E. \approx (8.0 \times 10^{-19}) - (4.8 \times 10^{-19})\,\text{J} = 3.2 \times 10^{-19}\,\text{J}.
Step 5: Calculate the Electron’s Momentum
The momentum of the electron can be found using
p = \sqrt{2 m_e \times K.E.}.
Substituting the values:
p = \sqrt{2 \times (9.1 \times 10^{-31}\,\text{kg}) \times (3.2 \times 10^{-19}\,\text{J})}.
Let’s approximate inside the square root first:
2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}
= 58.24 \times 10^{-50}\,\text{(kg·J)}
= 5.824 \times 10^{-49}.
Taking the square root gives
p \approx \sqrt{5.824 \times 10^{-49}}
\approx 7.63 \times 10^{-25}\,\text{kg·m/s}.
Step 6: Use de Broglie Relation to Find the Wavelength
The de Broglie wavelength is given by
\lambda_{\text{de Broglie}} = \frac{h}{p}.
Substituting the values:
\lambda_{\text{de Broglie}}
= \frac{6.63 \times 10^{-34}\,\text{J·s}}{7.63 \times 10^{-25}\,\text{kg·m/s}}.
Dividing these, we get approximately
\lambda_{\text{de Broglie}} \approx 8.7 \times 10^{-10}\,\text{m}.
Converting to Angstroms ( 1\,\text{Å} = 10^{-10}\,\text{m} ),
\lambda_{\text{de Broglie}} \approx 8.7\,\text{Å}
\approx 9\,\text{Å}\;(\text{rounded to the nearest integer}).
Final Answer
The de-Broglie wavelength of the emitted electrons is
9\,\text{Å}\;(\text{approximately}).
