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\(^{-1}\)
Show 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!
Step 1: Understanding the Question:
This is a problem based on the photoelectric effect. We are given the threshold wavelength (\(\lambda_0 = 500\text{ nm}\)) and the incident light wavelength (\(\lambda = 300\text{ nm}\)). We need to calculate the total kinetic energy of all photoelectrons ejected per second when the metal is illuminated by a 100 Watt source. Step 2: Key Formula or Approach:
1. The energy of a single incident photon is:
\[ E = \frac{hc}{\lambda} \]
2. The threshold energy (work function, \(W_0\)) of the metal is:
\[ W_0 = \frac{hc}{\lambda_0} \]
3. The kinetic energy of a single ejected photoelectron is:
\[ KE_{\text{single}} = E - W_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]
4. The power of the bulb is \(P = 100\text{ W} = 100\text{ J/s}\). This is the total incident energy per second.
5. The number of photons falling on the metal per second is:
\[ N = \frac{P}{E} \]
6. Assuming every photon with \(E > W_0\) ejects one photoelectron, the total kinetic energy of all photoelectrons ejected per second is:
\[ KE_{\text{total}} = N \times KE_{\text{single}} = \frac{P}{E} \times (E - W_0) = P \left( 1 - \frac{W_0}{E} \right) \] Step 3: Detailed Explanation:
Let's simplify the ratio of work function to photon energy:
\[ \frac{W_0}{E} = \frac{\frac{hc}{\lambda_0}}{\frac{hc}{\lambda}} = \frac{\lambda}{\lambda_0} \]
Substitute the given wavelengths into this ratio:
\[ \frac{W_0}{E} = \frac{300\text{ nm}}{500\text{ nm}} = 0.6 \]
Now, substitute this ratio into our formula for total kinetic energy per second:
\[ KE_{\text{total}} = P \left( 1 - \frac{\lambda}{\lambda_0} \right) \]
Given the power \(P = 100\text{ W}\):
\[ KE_{\text{total}} = 100\text{ J/s} \times (1 - 0.6) \]
\[ KE_{\text{total}} = 100 \times 0.4 = 40\text{ J/s} \]
Thus, the total kinetic energy of all photoelectrons ejected in one second is 40 Joules. Step 4: Final Answer:
The total kinetic energy of the ejected photoelectrons per second is 40 J, which corresponds to option (A).
