Problem 41 - Master Resource Book in JEE Main Physics

Question

Light of wavelength \(\lambda\) falls on a metal having work function \(\frac{h c}{\lambda_{0}} .\) Photoelectric effect will take place only if [DCE 2009] (a) \(\lambda \geq \lambda_{0}\) (b) \(\lambda \geq 2 \lambda_{0}\) (c) \(\lambda \leq \lambda_{0}\) (d) \(\lambda=4 \lambda_{0}\)

Step-by-Step Solution

Answer

(c) \(\lambda \leq \lambda_{0}\)

Step 1: Understand the Photoelectric Effect

The photoelectric effect occurs when photons strike a metal surface and eject electrons. For this to happen, the energy of the incident photon must be greater than the work function of the metal.

Step 2: Identify the Relation Between Wavelength and Photon Energy

The energy of a photon is given by the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light.

Step 3: Analyze the Given Conditions

The work function is given as \(\frac{hc}{\lambda_{0}}\). For the photoelectric effect to take place, the photon energy must be equal to or greater than the work function, so \(\frac{hc}{\lambda} \geq \frac{hc}{\lambda_{0}}\).

Step 4: Simplify the Inequality

Simplifying the inequality \(\frac{hc}{\lambda} \geq \frac{hc}{\lambda_{0}}\), we can cancel \(hc\) from both sides since they are constants, resulting in \(\frac{1}{\lambda} \geq \frac{1}{\lambda_{0}}\).

Step 5: Solve for Wavelength

Rearranging the inequality \(\frac{1}{\lambda} \geq \frac{1}{\lambda_{0}}\) gives \(\lambda \leq \lambda_{0}\). This means the wavelength of the incident light must be less than or equal to \(\lambda_{0}\) for the photoelectric effect to occur.

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon, the fundamental particle of light. To understand how photons can cause the photoelectric effect, we need to know about their energy first. Photon energy can be calculated using the equation \( E = \frac{hc}{\lambda} \). Here, \( h \) is Planck's constant, a tiny number that plays a crucial role in quantum mechanics, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.
This means that each photon has an energy inversely proportional to its wavelength - the shorter the wavelength, the higher the energy of the photon. When photons hit a surface like metal, their energy needs to be sufficiently high to eject electrons in the photoelectric effect.
To decide if a photon can eject an electron, we compare its energy to the material’s work function. If the photon's energy is greater than the work function, it can liberate an electron from the surface.

Work Function

The work function is an essential concept when discussing the photoelectric effect. It is defined as the minimum energy required to remove an electron from the surface of a metal. In simple terms, it's the energy "barrier" that a photon must overcome to eject an electron.
The dark-colored metals often have lower work functions, meaning that they can more easily release electrons when struck by photons of sufficient energy. In equations, we represent work function by \( \phi \), and in this exercise, it is given as \( \frac{hc}{\lambda_{0}} \), where \( \lambda_{0} \) is a specific threshold wavelength.
During the photoelectric effect, we relate the photon's energy to the work function. Electrons are ejected only if the photon energy \( E = \frac{hc}{\lambda} \) is equal to or exceeds this work function. Understanding the work function helps us predict which wavelengths of light can cause the photoelectric effect.

Wavelength

Wavelength is the distance between consecutive peaks of a wave. It plays a significant role in determining the energy of a photon. The relation between wavelength and photon energy is captured by the equation \( E = \frac{hc}{\lambda} \).
This means that a shorter wavelength corresponds to higher photon energy, which is crucial for understanding the photoelectric effect. In this context, if the wavelength \( \lambda \) is less than or equal to \( \lambda_{0} \), the threshold wavelength, the photons will have enough energy to overcome the work function and eject electrons from the metal.
A key feature of the photoelectric effect is that it only occurs for light with wavelengths shorter than this threshold value, which translates to having enough energy per photon to liberate electrons.