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Step-by-Step Solution
Step 1: Identify the Key Quantities
• Work function of sodium, W = 4.41 \times 10^{-19}\,\text{J}
• Wavelength of incident photon, \lambda = 300 \,\text{nm} = 300 \times 10^{-9}\,\text{m}
• Planck’s constant, h = 6.63 \times 10^{-34}\,\text{J}\,\text{s}
• Speed of light, c = 3 \times 10^{8}\,\text{m/s}
Step 2: Calculate the Energy of the Incident Photon
The energy E of a photon with wavelength \lambda is given by:
E = \frac{hc}{\lambda}
Substituting the known values:
E = \frac{(6.63 \times 10^{-34}\,\text{J}\,\text{s}) \times (3 \times 10^{8}\,\text{m/s})}{300 \times 10^{-9}\,\text{m}}
Simplify:
E = \frac{1.989 \times 10^{-25}\,\text{J}\,\text{m}}{300 \times 10^{-9}\,\text{m}}
= \frac{1.989 \times 10^{-25}}{3 \times 10^{-7}}\,\text{J}
= 6.63 \times 10^{-19}\,\text{J}
Step 3: Compute the Maximum Kinetic Energy of the Ejected Electrons
The maximum kinetic energy KE_{\text{max}} of the photoelectrons is:
KE_{\text{max}} = E - W
Substituting the values:
KE_{\text{max}} = 6.63 \times 10^{-19}\,\text{J} - 4.41 \times 10^{-19}\,\text{J}
= 2.22 \times 10^{-19}\,\text{J}
Step 4: Express the Answer in the Required Form
The problem requests the kinetic energy in units of 10^{-21}\,\text{J} . Thus:
2.22 \times 10^{-19}\,\text{J} = 222 \times 10^{-21}\,\text{J}
Therefore, the kinetic energy of the ejected electrons is 222 × 10–21 J.
