Question

Photoelectric emission is observed from a metal surface for frequencies \(v_{1}\) and \(v_{2}\) of the incident radiation \(\left(v_{1}>v_{2}\right)\). If maximum kinetic energies of the photoelectrons in the two cases are in the ratio \(1: K\), then the threshold frequency for the metal is given by (a) \(\frac{v_{2}-v_{1}}{K-1}\) (b) \(\frac{K v_{2}-v_{1}}{K-1}\) (c) \(\frac{K v_{1}-v_{2}}{K}\) (d) \(\frac{K v_{1}-v_{2}}{K-1}\)

Short answer

The threshold frequency for the metal is given by (b) \(\frac{K v_{2}-v_{1}}{K-1}\).

Step-by-step solution

Understand the Photoelectric Effect

The photoelectric effect is described by the equation: \(KE_{max} = hu - \phi\), where \(KE_{max}\) is the maximum kinetic energy of the photoelectrons, \(h\) is the Planck constant, \(u\) is the frequency of the incident light, and \(\phi\) is the work function of the metal or the threshold energy. The threshold frequency \(u_0\) is the frequency at which \(KE_{max}\) becomes zero, and it is related to the work function by \(\phi = hu_0\).

Apply the Photoelectric Equation to Both Frequencies

Set up the photoelectric equation for both frequencies \(u_1\) and \(u_2\):\[KE_{max_1} = hu_1 - \phi\] and \[KE_{max_2} = hu_2 - \phi\]. Since \(\phi = hu_0\), we can also express these equations in terms of the threshold frequency. The problem states that the maximum kinetic energies are in the ratio \(1:K\), i.e., \(KE_{max_2} = K \times KE_{max_1}\).

Express the Ratios of Kinetic Energies and Substitute the Threshold Frequency

Using the information given about the energies, we can write: \[K \times (hu_1 - \phi) = hu_2 - \phi\]. Substituting \(\phi = hu_0\) in this equation, we get: \[K \times (hu_1 - hu_0) = hu_2 - hu_0\]. Simplifying this, we have \[Khu_1 - Khu_0 = hu_2 - hu_0\].

Rearrange the Equation to Solve for the Threshold Frequency

Rearrange the above equation to isolate \(hu_0\):\[Khu_0 - hu_0 = hu_2 - Khu_1\]. Factor out \(h\) and \(u_0\) from the left side: \[(K - 1)hu_0 = hu_2 - Khu_1\]. Simplify by dividing by \(h(K - 1)\) to get: \[u_0 = \frac{u_2 - Ku_1}{K - 1}\].

Choose the Correct Option

By matching the result of Step 4 with the given options, we conclude that the correct threshold frequency for the metal is given by (b) \(\frac{K v_{2}-v_{1}}{K-1}\).

Key concepts

Threshold Frequency

When discussing the photoelectric effect, a foundational concept is the threshold frequency. This is simply the minimum frequency of incident light necessary to eject electrons from a metal surface. It's crucial because if the frequency of the incident light is below this threshold, no photoelectric emission will occur, regardless of the light's intensity.

Imagine the threshold frequency as a gatekeeper, determining whether or not the electromagnetic energy is sufficient to allow an electron to escape from the surface of the metal. This frequency is directly related to the unique property of the metal known as the work function, which quantifies the minimum energy required to remove an electron from the metal's surface.

By using the equation \(\phi = h u_0\) where \(\phi\) is the work function, \(h\) is Planck's constant and \(u_0\) is the threshold frequency, we can see how these concepts intertwine. A robust understanding of this relationship will greatly enhance a student's ability to tackle various photoelectric effect problems.

Photoelectric Emission

Photoelectric emission is fascinating—the way a metal surface can emit electrons when it's struck by light. It's essential to understand that this emission is not a continuous process but happens only when certain conditions are met. Among these conditions, as mentioned earlier, is that the incident light must surpass the threshold frequency, but it's also imperative that the light deliver enough energy to overcome the electrons' binding energy in the metal.

Once the light's energy is high enough, we see this remarkable phenomena in action: electrons, or photoelectrons, get ejected. They exit with varying amounts of kinetic energy, depending on the energy of the incoming photons minus the work function of the metal. The beautiful simplicity of Albert Einstein's explanation of this process, which ultimately earned him a Nobel Prize, lies in the direct proportionality between the light's frequency and the maximum kinetic energy of the ejected photoelectrons. This is perfectly encapsulated in the equation: \(KE_{max} = h(u - u_0)\).

It is this concept that's at play in the original exercise, where students are tasked to understand and manipulate the relationships between incident light frequencies and the resulting kinetic energies of the photoelectrons.

Maximum Kinetic Energy of Photoelectrons

Now let's dive into the concept of maximum kinetic energy of photoelectrons. This is essentially the highest amount of kinetic energy that a photoelectron can achieve when ejected by photoelectric emission. It can be thought of as a measure of the 'excess' energy the electron has after freeing itself from the metal.

This maximum kinetic energy is pivotal in calculating everything from the threshold frequency to the speed of photoelectrons. When we write down the fundamental equation \(KE_{max} = h(u - u_0)\), which suggests that \(KE_{max}\) depends linearly on the frequency \(u\) of incident light, we're observing a key aspect of the photoelectric effect: varying the frequency (assuming it's above the threshold frequency) can increase or decrease the \(KE_{max}\) of photoelectrons linearly. Recognizing the linear relationship between the frequency of incident light and the maximum kinetic energy helps students predict how changes in the incident light's frequency will affect the energy transfer process in photoelectric emission.

Understanding this maximum kinetic energy is also crucial in interpreting our original problem, where we're given the ratio of kinetic energies resulting from different incident light frequencies, and asked to deduce the threshold frequency. Clearly, grasping this concept is vital for students looking to excel in understanding photoelectric effect scenarios.