Question:medium

Light of wavelength ' $\lambda$ ' falls on a metal having work function $\frac{\text{hc}{\lambda_0}$. Photoelectric effect will take place only if ( $\lambda_0$ is the threshold wavelength)}

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Higher Frequency = More Energy, but Higher Wavelength = Less Energy. To eject electrons, you need "Short" wavelengths or "High" frequencies.
Updated On: May 14, 2026
  • $\lambda \geq \lambda_0$
  • $\lambda \geq 2\lambda_0$
  • $\lambda \leq \lambda_0$
  • $\lambda = 4\lambda_0$
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The photoelectric effect occurs only when the energy of the incident light is greater than or equal to the work function of the metal. The work function is the minimum energy required to eject an electron from the metal surface.
Step 2: Key Formula or Approach:
The energy $E$ of an incident photon is given by $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.
The work function $\Phi$ is given as $\frac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.
The condition for the photoelectric effect is: $E \geq \Phi$.
Step 3: Detailed Explanation:
Substitute the expressions for $E$ and $\Phi$ into the condition:
\[ \frac{hc}{\lambda} \geq \frac{hc}{\lambda_0} \]
Since $h$ and $c$ are positive constants, we can cancel them from both sides:
\[ \frac{1}{\lambda} \geq \frac{1}{\lambda_0} \]
Taking the reciprocal of both sides reverses the inequality sign (since both wavelengths are positive quantities):
\[ \lambda \leq \lambda_0 \]
Step 4: Final Answer:
The photoelectric effect will take place only if $\lambda \leq \lambda_0$.
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