Question:

Photoemission of electrons occurs from a metal \((\phi_0=1.96\,\text{eV})\) when light of frequency \(6.4\times10^{14}\,\text{Hz}\) is incident on it. Calculate: Energy of a photon in the incident light.

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For photoelectric-effect problems, it is often convenient to express photon energy directly in electron volts: \[ E=h\nu. \] After finding energy in joules, divide by \[ 1.6\times10^{-19} \] to convert it into eV.
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Solution and Explanation

Concept: According to Planck's quantum theory, light consists of packets of energy called photons. The energy carried by each photon is directly proportional to the frequency of the incident radiation and is given by Planck's relation \[ E=h\nu, \] where
• \(E\) = energy of one photon,
• \(h=6.626\times10^{-34}\,\text{J s}\) = Planck's constant,
• \(\nu\) = frequency of incident radiation. Thus, to determine the energy of the incident photon, we simply multiply Planck's constant by the given frequency.

Step 1:
Write the given quantities. Frequency of incident light: \[ \nu=6.4\times10^{14}\,\text{Hz} \] Planck's constant: \[ h=6.626\times10^{-34}\,\text{J s} \]

Step 2:
Apply Planck's equation for photon energy. The energy of one photon is \[ E=h\nu. \] Substituting the given values, \[ E = (6.626\times10^{-34}) (6.4\times10^{14}). \]

Step 3:
Perform the numerical calculation carefully. Multiplying the numerical coefficients, \[ 6.626\times6.4 = 42.4064. \] Therefore, \[ E = 42.4064\times10^{-20}\,\text{J}. \] Writing in standard scientific notation, \[ E = 4.24064\times10^{-19}\,\text{J}. \] Hence, \[ \boxed{ E \approx 4.24\times10^{-19}\,\text{J} }. \]

Step 4:
Convert the energy into electron volt (optional but useful in photoelectric problems). Since \[ 1\,\text{eV} = 1.6\times10^{-19}\,\text{J}, \] \[ E = \frac{4.24\times10^{-19}} {1.6\times10^{-19}} \,\text{eV}. \] \[ E = 2.65\,\text{eV}. \] Thus, \[ \boxed{ E \approx 2.65\,\text{eV} }. \] Final Answer: Energy of the incident photon \[ \boxed{ E=4.24\times10^{-19}\,\text{J} } \] or equivalently \[ \boxed{ E\approx2.65\,\text{eV}. } \]
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