Question:
In a photoelectric effect, the kinetic energy of electrons is \(1.3 \times 10^{-19}\,\text{J}\). If work function of the metal is \(2.27\,\text{eV}\), then the frequency (in Hz) of the incident radiation is (\(h = 6.6 \times 10^{-34}\,\text{J s}\), \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\)).
In a photoelectric effect, the kinetic energy of electrons is \(1.3 \times 10^{-19}\,\text{J}\). If work function of the metal is \(2.27\,\text{eV}\), then the frequency (in Hz) of the incident radiation is (\(h = 6.6 \times 10^{-34}\,\text{J s}\), \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\)).
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For photoelectric effect:
\[
hf=\phi+K_{\max}
\]
Always convert the work function from eV to joules before substituting into the equation.
Updated On: Jul 9, 2026
- \(7.47 \times 10^{14}\)
- \(7.47 \times 10^{15}\)
- \(6.47 \times 10^{15}\)
- \(3.47 \times 10^{14}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
According to Einstein's photoelectric equation,
\[
hf=\phi+K_{\max}
\]
where
• \(h\) = Planck's constant,
• \(\phi\) = work function,
• \(K_{\max}\) = maximum kinetic energy.
Step 1: Convert the work function into joules. \[ \phi=2.27\times1.6\times10^{-19} =3.632\times10^{-19}\,\text{J} \]
Step 2: Find the energy of the incident photon. \[ E=\phi+K_{\max} =(3.632+1.3)\times10^{-19} =4.932\times10^{-19}\,\text{J} \]
Step 3: Calculate the frequency. \[ f=\frac{E}{h} =\frac{4.932\times10^{-19}}{6.6\times10^{-34}} =7.47\times10^{14}\,\text{Hz} \]
Step 4: Final conclusion. \[ \boxed{f=7.47\times10^{14}\,\text{Hz}} \] Hence, the correct option is \(\boxed{(B)}\).
• \(h\) = Planck's constant,
• \(\phi\) = work function,
• \(K_{\max}\) = maximum kinetic energy.
Step 1: Convert the work function into joules. \[ \phi=2.27\times1.6\times10^{-19} =3.632\times10^{-19}\,\text{J} \]
Step 2: Find the energy of the incident photon. \[ E=\phi+K_{\max} =(3.632+1.3)\times10^{-19} =4.932\times10^{-19}\,\text{J} \]
Step 3: Calculate the frequency. \[ f=\frac{E}{h} =\frac{4.932\times10^{-19}}{6.6\times10^{-34}} =7.47\times10^{14}\,\text{Hz} \]
Step 4: Final conclusion. \[ \boxed{f=7.47\times10^{14}\,\text{Hz}} \] Hence, the correct option is \(\boxed{(B)}\).
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