Question

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and the stopping potential is 2V, what is the wavelength of the electromagnetic wave? Given \( hc = 1242 \, \text{eV} \cdot \text{nm} \) where \( h \) is the Planck constant and \( c \) is the speed of light in vacuum.

• A. 400 nm
• B. 600 nm
• C. 200 nm
• D. 300 nm

Hint:

The energy of the photon in the photoelectric effect can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.

Answer 1

The Correct Option is D

The problem given is based on the photoelectric effect, a phenomenon where light incident on a metal surface causes the emission of electrons. We are provided with the work function of the metal and the stopping potential, both of which are essential in determining the wavelength of the incident electromagnetic wave.

Given: 

• Work function (\( \phi \)) = 2.14 eV
• Stopping potential (\( V_0 \)) = 2 V
• Product of Planck's constant and speed of light (\( hc \)) = 1242 eV·nm

To find the wavelength (\( \lambda \)) of the electromagnetic wave, we need to follow these steps:

The total energy (\( E \)) of the incident photon is given by the sum of the work function and the energy required to overcome the stopping potential:

1. \(E = \phi + eV_0 = 2.14 \, \text{eV} + 2 \, \text{eV} = 4.14 \, \text{eV}\)

The energy (\( E \)) of the incident photon can also be expressed in terms of its wavelength (\( \lambda \)) using the relation:

1. \(E = \frac{hc}{\lambda}\)

Substitute the known values and solve for \( \lambda \):

1. \(4.14 = \frac{1242}{\lambda}\)

Rearrange the equation to solve for \( \lambda \):

1. \(\lambda = \frac{1242}{4.14} \approx 300 \, \text{nm}\)

Thus, the wavelength of the electromagnetic wave is 300 nm.

Let's verify the options given:

• 400 nm
• 600 nm
• 200 nm
• 300 nm

The correct answer from the options is indeed 300 nm.

Answer 2

The energy of the incident photon is given by: \[ E_{\text{photon}} = \frac{hc}{\lambda}. \] The total energy of the photon is used to overcome the work function \( \phi \) of the metal and to provide the kinetic energy to the ejected electrons. Thus: \[ E_{\text{photon}} = \phi + K.E. \] Substitute the values: \[ \frac{hc}{\lambda} = 2.14 \, \text{eV} + 2 \, \text{eV} = 4.14 \, \text{eV}. \] Using \( hc = 1242 \, \text{eV} \cdot \text{nm} \), we get: \[ \frac{1242}{\lambda} = 4.14 \quad \Rightarrow \quad \lambda = \frac{1242}{4.14} \approx 300 \, \text{nm}. \]