Question

The minimum energy needed to remove an electron from a metal corresponds to a wavelength of 500 nm. What is the total kinetic energy of all the photoelectrons ejected per second when the entire radiation from a 100 Watt bulb with a wavelength of 300 nm falls on the surface of the metal?
Planck’s constant = 6.6 \(\times 10^{-34}\) J s; speed of light = 3 \(\times 10^8\) m s\(^{-1}\)

• A. 40 J
• B. 2.6 \(\times 10^{-19}\) J
• C. 1.6 \(\times 10^{-19}\) J
• D. 80 J

Hint:

Using the ratio \(\frac{KE_{\text{total}}}{P} = 1 - \frac{\lambda}{\lambda_0}\) bypasses all tedious calculations involving Planck's constant and the speed of light.
This shortcut is extremely powerful and saves invaluable time during exams!

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the total kinetic energy of all photoelectrons ejected per second when the entire radiation from the 100 Watt bulb falls on the metal surface. This involves understanding the photoelectric effect.

1. First, we calculate the energy of the photons corresponding to the wavelength that can just remove an electron (threshold energy).
   • The wavelength given for the minimum energy is 500 nm. Convert this to meters: \(500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\).
   • We use the formula: \(E = \frac{hc}{\lambda}\) to find the energy. Here, \(h = 6.6 \times 10^{-34} \, \text{J s}\) and \(c = 3 \times 10^8 \, \text{m/s}\).
   • Calculate the energy: \(E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}}\)
   • This gives \(E = 3.96 \times 10^{-19} \, \text{J}\), which is the work function \((\Phi)\).
2. Next, calculate the energy of photons with a wavelength of 300 nm (which is given for the bulb radiation).
   • Convert the wavelength to meters: \(300 \, \text{nm} = 300 \times 10^{-9} \, \text{m}\).
   • Again, use the energy formula: \(E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}}\).
   • This results in photon energy \(E \approx 6.6 \times 10^{-19} \, \text{J}\).
3. Now, find the kinetic energy of a single photoelectron ejected, which is the difference.
   • The kinetic energy \((K.E.)\) of one photoelectron is given by: \(K.E. = h\nu - \Phi\).
   • So, \(K.E. = 6.6 \times 10^{-19} - 3.96 \times 10^{-19} = 2.64 \times 10^{-19} \, \text{J}\).
4. Finally, use the power of the bulb to find the number of photons and the total kinetic energy per second.
   • The power of the bulb is 100 W, which means 100 Joules per second.
   • The number of photons emitted per second can be found using: \(n = \frac{\text{Power}}{\text{Photon energy}} = \frac{100}{6.6 \times 10^{-19}}\).
   • The number of photons is approximately \(1.515 \times 10^{20} \, \text{photons/s}\).
   • Thus, the total kinetic energy of all ejected photoelectrons per second is: \(\text{Total K.E.} = n \times (K.E. \text{ of one electron})\)
   • This gives: \(\text{Total K.E.} = 1.515 \times 10^{20} \times 2.64 \times 10^{-19} \approx 40 \, \text{J}\).

Therefore, the total kinetic energy of all the photoelectrons ejected per second is 40 J.