---
title: "Monochromatic light of wavelength \\(\\lambda_1\\) is incident on a metal surface in a photoelectric experiment, resulting in photoelectrons emitted with a stopping potential of \\(V_1\\). The light source is then replaced with another monochromatic source emitting light of wavelength \\(\\lambda_2 = \\dfrac{\\lambda_1}{2}\\), resulting in a stopping potential of \\(V_2\\). Which of the following correctly compares \\(V_2\\) to \\(V_1\\) and provides a valid justification?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/116515/"
date_modified: "2026-08-04T05:02:21+00:00"
---

# Monochromatic light of wavelength \(\lambda_1\) is incident on a metal surface in a photoelectric experiment, resulting in photoelectrons emitted with a stopping potential of \(V_1\). The light source is then replaced with another monochromatic source emitting light of wavelength \(\lambda_2 = \dfrac{\lambda_1}{2}\), resulting in a stopping potential of \(V_2\). Which of the following correctly compares \(V_2\) to \(V_1\) and provides a valid justification?

Monochromatic light of wavelength \(\lambda_1\) is incident on a metal surface in a photoelectric experiment, resulting in photoelectrons emitted with a stopping potential of \(V_1\). The light source is then replaced with another monochromatic source emitting light of wavelength \(\lambda_2 = \dfrac{\lambda_1}{2}\), resulting in a stopping potential of \(V_2\). Which of the following correctly compares \(V_2\) to \(V_1\) and provides a valid justification?

- **A.** \(V_2 = 2V_1\), because halving the incident wavelength doubles the photon energy, which doubles the maximum kinetic energy of the emitted photoelectrons.
- **B.** \(V_2 < 2V_1\), because the work function of the metal surface increases as the energy of the incident photons increases.
- **C.** \(V_2 > 2V_1\), because halving the wavelength doubles the photon energy, and subtracting the constant work function from a doubled photon energy yields more than double the maximum kinetic energy.
- **D.** \(V_2 = 4V_1\), because stopping potential is proportional to the square of the frequency of the incident electromagnetic wave.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/116515/*