Question

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)}

• A. $\lambda \geq \lambda_0$
• B. $\lambda \geq 2\lambda_0$
• C. $\lambda \leq \lambda_0$
• D. $\lambda = 4\lambda_0$

Hint:

Higher Frequency = More Energy, but Higher Wavelength = Less Energy. To eject electrons, you need "Short" wavelengths or "High" frequencies.

Answer 1

The Correct Option is C

Step 1: Concept
For the photoelectric effect to occur, the energy of the incident photon ($E$) must be greater than or equal to the work function ($\Phi$) of the metal.

Step 2: Meaning
Energy is given by $E = \frac{hc}{\lambda}$ and the work function is $\Phi = \frac{hc}{\lambda_0}$.

Step 3: Analysis
The condition $E \geq \Phi$ becomes $\frac{hc}{\lambda} \geq \frac{hc}{\lambda_0}$. Since $\lambda$ is in the denominator, this inequality implies $\lambda \leq \lambda_0$.

Step 4: Conclusion
Photoelectric emission only happens when the incident wavelength is less than or equal to the threshold wavelength. Final Answer: (C)