Electron Ejection and Photon Energy

The video tutorial explains how to calculate the velocity of electrons ejected from cesium metal when exposed to blue light with a wavelength of 475 nanometers. It covers the calculation of the light's frequency, the energy of photons, and the work function of cesium. The tutorial then determines the kinetic energy of ejected electrons and calculates their velocity, ensuring the results are reasonable compared to the speed of light.

What is the first step in calculating the velocity of ejected electrons when blue light is shone on cesium metal?

• Measure the wavelength of the light

• Find the mass of the electron

• Determine the frequency of the blue light

• Calculate the speed of light

How is the energy of a photon related to its frequency?

• Energy is equal to the square of the frequency

• Energy is directly proportional to frequency

• Energy is independent of frequency

• Energy is inversely proportional to frequency

What is the work function of cesium metal in electron volts?

• 2.1 eV

• 1.5 eV

• 4.2 eV

• 3.0 eV

Why is it necessary to convert the work function from electron volts to joules?

• To compare with the speed of light

• To match the units of Planck's constant

• To simplify the calculation

• Because joules are the SI unit of energy

What must be true for electrons to be ejected from cesium metal?

• The energy of the photon must be zero

• The energy of the photon must be greater than the work function

• The energy of the photon must be equal to the work function

• The energy of the photon must be less than the work function

What is the kinetic energy of the ejected electrons if the energy of the photon is greater than the work function?

• Equal to the work function

• The difference between the photon energy and the work function

• Equal to the photon energy

• The sum of the photon energy and the work function

What is the mass of an electron used in the velocity calculation?

• 6.63 x 10^-34 kg

• 9.11 x 10^-31 kg

• 1.67 x 10^-27 kg

• 1.60 x 10^-19 kg

Why is it important to check if the calculated velocity is a significant fraction of the speed of light?

• To confirm the wavelength of the light

• To verify the mass of the electron

• To determine if relativistic effects are important

• To ensure the calculation is correct

What is the final calculated velocity of the ejected electrons?

• 4.18 x 10^5 m/s

• 1.50 x 10^6 m/s

• 2.99 x 10^8 m/s

• 3.85 x 10^5 m/s

What would happen if the velocity of the ejected electrons was a significant fraction of the speed of light?

• The work function would change

• Relativistic effects would need to be considered

• The calculation would be incorrect

• The electrons would not be ejected