Pieces of potassium (Φ = 3.68 × 10–19 J) and sodium (Φ = 4.41 × 10–19 J) metal are exposed to radiation of wavelength 290.0 nm.

Question

What is the velocity of the electrons from sodium?

What is the velocity of the electrons from potassium?

Answer 1

Hi Esha K.!

Step 1: Energy of the Photon

We must first use the Planck relation (or Planck-Einstein relation) to determine the energy of the photon. We use the following equation:

E = hf

However, we were not given the frequency of the photon. So, we must use the wave equation as well:

c = λf => f = c*λ^-1

Now, we will combine these two equations and solve:

E = hf

=> c = λf => f = c*λ^-1

E = h(c*λ^-1)

<=> [(6.626x10^-34 J*s)*(2.99x10⁸ m*s^-1)]/(290x10^-9 m)

E_photon = 6.83x10^-19 J

Step 2: The Photoelectric Effect

There is much to say regarding this facet of the solution when it comes to theory. Suffice it to say, Einstein won the 1921 Nobel Prize in physics for his work on the photoelectric effect.

First, we will make use of his equation:

E_photon = KE_electron + Φ

KE_electron = E_photon - Φ

Next, we will substitute in the following equation for kinetic energy:

KE = (1/2)mv²

We will then rearrange:

KE_electron = E_photon - Φ

=> KE = (1/2)mv²

(1/2)mv² = E_photon - Φ => v = √[2(E_photon - Φ)/m_electron]

Step 3: Solve

We will now use the specific values for the work function for each element and solve for the velocity of the ejected electron:

v_K = √[2(E_photon - Φ_K)/m_electron]

<=> √[2(6.83x10^-19 J - 3.68x10^-19 J)/9.11x10^-31 kg]

v_K = 8.32x10⁵ m*s^-1

v_Na = √[2(E_photon - Φ_Na)/m_electron]

<=> √[2(6.83x10^-19 J - 4.41x10^-19 J)/9.11x10^-31 kg]

v_Na = 7.29x10⁵ m*s^-1

Hope this helps!

Cheers