A source emits monochromatic light of frequency `5.5xx10^(14)` Hzat a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of `6muA` in an extrnal circuit.
(a) Find the enrgy of a photon.
(b) Find the number of photons leaving the source per second.
(C) Find the percentage of the photons falling on the cathode which produce photoelectrons.

Explanation

Let's solve the problem step by step.

### Given Data:
- Frequency of light, \( f = 5.5 \times 10^{14} \) Hz
- Power of the source, \( P = 0.1 \) W
- Percentage of photons falling on the cathode, \( 0.15\% \)
- Current produced in the circuit, \( I = 6 \, \mu A = 6 \times 10^{-6} \, A \)

### (a) Find the energy of a photon.

The energy of a photon can be calculated using the formula:
\[ E = h \cdot f \]
where \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, J \cdot s \).

Substituting the values:
\[ E = (6.63 \times 10^{-34} \, J \cdot s) \cdot (5.5 \times 10^{14} \, Hz) \]
Calculating this gives:
\[ E = 3.6415 \times 10^{-19} \, J \]
To convert this energy into electron volts (eV), we use the conversion \( 1 \, eV = 1.6 \times 10^{-19} \, J \):
\[ E = \frac{3.6415 \times 10^{-19} \, J}{1.6 \times 10^{-19} \, J/eV} \approx 2.276 \, eV \]

### (b) Find the number of photons leaving the source per second.

The number of photons emitted per second, \( n_1 \), can be calculated using the formula:
\[ n_1 = \frac{P}{E} \]
Substituting the values:
\[ n_1 = \frac{0.1 \, W}{3.6415 \times 10^{-19} \, J} \]
Calculating this gives:
\[ n_1 \approx 2.747 \times 10^{17} \, photons/s \]

### (c) Find the percentage of the photons falling on the cathode which produce photoelectrons.

First, we need to find the number of photons falling on the cathode, \( n_2 \):
\[ n_2 = 0.15\% \text{ of } n_1 = \frac{0.15}{100} \cdot n_1 \]
Calculating \( n_2 \):
\[ n_2 = 0.0015 \cdot (2.747 \times 10^{17}) \approx 4.121 \times 10^{14} \, photons \]

Next, we find the number of photoelectrons produced, \( n_p \):
\[ n_p = \frac{I}{e} \]
where \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \, C \):
\[ n_p = \frac{6 \times 10^{-6} \, A}{1.6 \times 10^{-19} \, C} \approx 3.75 \times 10^{13} \, electrons \]

Finally, we calculate the percentage of photons falling on the cathode that produce photoelectrons:
\[ \text{Percentage} = \left( \frac{n_p}{n_2} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{3.75 \times 10^{13}}{4.121 \times 10^{14}} \right) \times 100 \approx 9.09\% \]

### Summary of Results:
(a) Energy of a photon: \( \approx 2.276 \, eV \)
(b) Number of photons leaving the source per second: \( \approx 2.747 \times 10^{17} \, photons/s \)
(c) Percentage of photons falling on the cathode which produce photoelectrons: \( \approx 9.09\% \)