Question

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

• A. Green
• B. Blue
• C. Red
• D. Yellow

Hint:

Use the photoelectric equation to relate the energy of incident light to the work function and kinetic energy of emitted electrons. Remember that shorter wavelengths correspond to higher energy photons, which are required to overcome the work function.

Answer 1

The Correct Option is B

To determine the color of light required to cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to understand the concept of the photoelectric effect. In this phenomenon, electrons are emitted from a metal surface when light of sufficient energy strikes it.

The energy \(E\) required to eject an electron is given by the equation:

\(E = h \cdot f\)

Where:

• \(E\) is the energy of the incident light (must be equal to or greater than the work function),
• \(h\) is Planck's constant \((4.135667696 \times 10^{-15} \text{ eV} \cdot \text{s})\), and
• \(f\) is the frequency of the light.

The work function \(W_0\) of the metal is the minimum energy required to remove an electron from the metal surface. For this problem:

\(W_0 = 3 \, \text{eV}\)

Since energy also relates to wavelength via:

\(E = \frac{h \cdot c}{\lambda}\)

Where:

• \(c\) is the speed of light \((3 \times 10^8 \, \text{m/s})\), and
• \(\lambda\) is the wavelength.

Given \(E = 3 \, \text{eV}\), we solve for the wavelength \(\lambda\):

\(\lambda = \frac{h \cdot c}{E} = \frac{4.135667696 \times 10^{-15} \times 3 \times 10^8}{3}\)

Converting \(\lambda\) into nanometers:

\(\lambda = \frac{(4.135667696 \times 3 \times 10^8)}{3} \times 10^9 \approx 414 \, \text{nm}\)

This wavelength corresponds to blue light in the visible spectrum, which ranges from about 450 nm to 495 nm. Thus, blue light is indeed adequate to cause the photoelectric emission for this metal.

Hence, the correct answer is: Blue.

Answer 2

The question requires determining the color of visible light necessary to cause the emission of photoelectrons from a metal with a work function of 3 eV. Let's solve this step by step:

Theoretical Background:

The work function, denoted by \(\phi\), is the minimum energy required to remove an electron from the surface of a metal. For the photoelectric effect to occur, the energy of the incident light must be equal to or greater than the work function. According to Einstein's photoelectric equation:

\(E = h\nu = \phi + KE_{max}\)

Where:

• \(E\) is the energy of the incident light.
• \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{Js}\)).
• \(\nu\) is the frequency of the incident light.
• \(KE_{max}\) is the maximum kinetic energy of emitted electrons (which is zero at the threshold).
• \(\phi = 3 \ \text{eV} = 3 \times 1.602 \times 10^{-19} \text{J}\)

Conversion and Calculation:

Since the work function is given, the threshold frequency \(\nu_0\) can be calculated as follows:

\(\nu_0 = \frac{\phi}{h}\)

Let's convert the work function:

\(\phi = 3 \ \text{eV} = 3 \times 1.602 \times 10^{-19} \text{J} = 4.806 \times 10^{-19} \text{J}\)

Calculate the threshold frequency:

\(\nu_0 = \frac{4.806 \times 10^{-19} \text{J}}{6.626 \times 10^{-34} \text{Js}} \approx 7.25 \times 10^{14} \ \text{Hz}\)

Determining the Wavelength:

The wavelength \(\lambda\) of the light can be found using the speed of light \(c = 3 \times 10^{8} \text{m/s}\) and the formula:

\(\lambda = \frac{c}{\nu_0} \approx \frac{3 \times 10^{8} \text{m/s}}{7.25 \times 10^{14} \ \text{Hz}} \approx 413 \ \text{nm}\)

Conclusion:

The calculated wavelength (approximately 413 nm) falls within the blue region of the visible spectrum, which ranges from about 450 nm to 495 nm. Hence, visible light with the color "Blue" is required to cause the emission of photoelectrons.

Correct Answer: Blue