Question

If \( \lambda \) is the incident wavelength and \( \lambda_0 \) is the threshold wavelength for a metal surface, photoelectric effect takes place only if:

• A. \( \lambda \le \lambda_0 \)
• B. \( \lambda \ge \lambda_0 \)
• C. \( \lambda \ge 2\lambda_0 \)
• D. None of these

Hint:

Shorter wavelength (higher frequency) is needed for photoelectric emission.

Answer 1

The Correct Option is A

The question pertains to the photoelectric effect, which is a phenomenon observed when light of a certain wavelength hits a metal surface, thereby causing the ejection of electrons. The wavelength of the incident light significantly affects whether the photoelectric effect will occur.

To understand the conditions for the photoelectric effect, we must consider the following:

• Threshold Wavelength (\(\lambda_0\)): This is the maximum wavelength of light that can cause the ejection of electrons from a metal surface. Light with a wavelength longer than \(\lambda_0\) will not have enough energy to cause the ejection of electrons.
• Incident Wavelength (\(\lambda\)): This is the wavelength of light that strikes the metal surface.

The key criterion for the photoelectric effect to take place is that the energy of the incident photons must be greater than or equal to the energy required to remove an electron from the metal surface. Energy of a photon, \(E\), is inversely proportional to its wavelength, as expressed by the equation:

\(E = \frac{hc}{\lambda}\)

where \(h\) is Planck's constant and \(c\) is the speed of light in vacuum. For the photoelectric effect to occur, the following condition must be met:

The energy of incident photons \( (\frac{hc}{\lambda}) \) should be greater than or equal to the work function \((W_0)\) of the metal surface, where \(W_0\) corresponds to the maximum wavelength or threshold condition.

This implies:

\(\frac{hc}{\lambda} \ge \frac{hc}{\lambda_0}\)

After simplifying, we get:

\(\lambda \le \lambda_0\)

Thus, for the photoelectric effect to take place, the wavelength of the incident light (\(\lambda\)) must be less than or equal to the threshold wavelength (\(\lambda_0\)). This means only shorter wavelengths (higher energy) than the threshold will suffice.

Conclusion: The correct answer is \( \lambda \le \lambda_0 \).