Problem 15 - Master Resource Book in JEE Main Physics

Question

The photoelectric threshold of Tungsten is \(2300 \AA\). The energy of the electrons ejected from the surface by ultraviolet light of wavelength \(1800 \mathrm{~A}\) is \(\left(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (a) \(0.15 \mathrm{eV}\) (b) \(1.5 \mathrm{eV}\) (c) \(15 \mathrm{eV}\) (d) \(150 \mathrm{eV}\)

Step-by-Step Solution

Answer

(b) 1.5 eV

Step 1: Understand the Problem

We need to find the energy of the ejected electrons when tungsten is illuminated by light with a given wavelength. The given data includes the threshold wavelength and the wavelength of the incident light. The work function can be calculated from the threshold wavelength, and the energy of the incident light can be calculated using its wavelength.

Step 2: Calculate Work Function

The work function is the minimum energy needed to eject electrons from the surface. It is calculated as \( \phi = \frac{hc}{\lambda_0} \), where \( \lambda_0 = 2300\, \text{Å} = 2300 \times 10^{-10}\, \text{m} \). So, \( \phi = \frac{6.6 \times 10^{-34}\, \text{J-s} \times 3 \times 10^8\, \text{m/s}}{2300 \times 10^{-10}\, \text{m}} \), which simplifies to \( \phi \approx 8.62 \times 10^{-19}\, \text{J} \). Converting to electron volts: \( 1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J} \), thus \( \phi \approx 5.38\, \text{eV} \).

Step 3: Calculate Photon Energy

The energy of the incoming photon with wavelength \( \lambda = 1800\, \text{Å} = 1800 \times 10^{-10}\, \text{m} \) is calculated as \( E = \frac{hc}{\lambda} \). Thus, \( E = \frac{6.6 \times 10^{-34}\, \text{J-s} \times 3 \times 10^8\, \text{m/s}}{1800 \times 10^{-10}\, \text{m}} \), which gives \( E \approx 1.10 \times 10^{-18}\, \text{J} \). Converting this to electron volts: \( E \approx 6.88\, \text{eV} \).

Step 4: Compute Ejected Electron's Energy

The kinetic energy of ejected electrons is given by \( K.E. = E - \phi \). Substituting the values, \( K.E. = 6.88\, \text{eV} - 5.38\, \text{eV} \), which results in \( K.E. = 1.5\, \text{eV} \).

Step 5: Select the Correct Answer

The calculated kinetic energy of the ejected electrons is \( 1.5\, \text{eV} \), which matches option (b). Therefore, the correct answer is (b) \(1.5\, \text{eV}\).

Key Concepts

Work Function

In the realm of the photoelectric effect, the work function is a critical concept. It defines the minimum amount of energy required to eject an electron from the surface of a material. Think of it as the energy barrier electrons need to jump over to escape the material.
The work function is represented by the symbol \( \phi \) and is determined using the threshold wavelength \( \lambda_0 \). The equation is:

• \( \phi = \frac{hc}{\lambda_0} \)

Here, \( h \) is Planck's constant \( (6.6 \times 10^{-34} \, \text{J-s}) \), and \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \).
For tungsten, the threshold wavelength given is \( 2300 \, \text{Å} \). Substituting these values can help calculate the work function in electron volts. This calculation shows how the work function depends on the physical properties of the material and is intrinsic to each material's surface.

Threshold Wavelength

The threshold wavelength \( (\lambda_0) \) is another intrinsic property of materials in the photoelectric effect. It represents the maximum wavelength of light that can still cause electrons to be emitted from the surface. Light with a longer wavelength doesn't have enough energy to overcome the work function.
Mathematically, it is connected to the work function by:

• \( \lambda_0 = \frac{hc}{\phi} \)

With this equation, you can see how smaller threshold wavelengths correspond to higher work functions. For tungsten, the threshold wavelength is given as \( 2300 \, \text{Å} \). This represents the minimum energy light needed to cause electron ejection. Any light with a wavelength shorter than this threshold can potentially cause the photoelectric effect.

Photon Energy

Photon energy is pivotal in understanding how light prompts the ejection of electrons. This energy is fundamentally determined by the wavelength of the incoming light. Shorter wavelengths equate to higher photon energy because energy is inversely proportional to wavelength.
The formula for photon energy \( (E) \) is:

• \( E = \frac{hc}{\lambda} \)

Where \( \lambda \) is the wavelength of the incident light. For a wavelength of \( 1800 \, \text{Å} \), the energy can be calculated, highlighting the photon's potential to eject electrons if it exceeds the work function.
The photoelectric effect hinges on having photon energy greater than the work function, allowing the emission of electrons since excess energy becomes electron kinetic energy.

Kinetic Energy of Electrons

When light hits a material and liberates an electron, the remaining energy after surmounting the work function becomes kinetic energy. This is what gives the ejected electron a velocity.
The kinetic energy \( (K.E.) \) of the ejected electrons can be calculated using:

• \( K.E. = E - \phi \)

Where \( E \) is the photon energy, and \( \phi \) is the work function. This calculation shows the leftover energy after the necessary amount has been used to release the electron.
In the tungsten example, light with a wavelength of \( 1800 \, \text{Å} \) is used, and it was found that the kinetic energy of the electrons was \( 1.5 \, \text{eV} \). This illustrates how energy not used in overcoming the work function translates into the motion of electrons, a vivid demonstration of energy conversion in the photoelectric effect.