Question

The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV, respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of:

• A. Li only
• B. Cs only
• C. Neither Cs nor Li
• D. Both Cs and Li

Hint:

For the photoelectric effect to occur, the energy of the incoming light must be greater than the work function of the material.

Answer 1

The Correct Option is B

The energy of the incident photon is: \[ E = \frac{1240}{\lambda} = \frac{1240}{550} \approx 2.25 \, \text{eV} \] For the photoelectric effect to occur, the energy of the photon must be greater than the work function of the metal. 

For Cs (work function = 1.9 eV): 

Since \( 2.25 \, \text{eV} > 1.9 \, \text{eV} \), photoelectric effect is possible for Cs. 

For Li (work function = 2.5 eV): 

Since \( 2.25 \, \text{eV} <2.5 \, \text{eV} \), photoelectric effect is not possible for Li. 

Thus, the answer is \( \boxed{\text{Cs only}} \).

Answer 2

Energy of incident photons: The energy of a photon is given by \[ E = \frac{hc}{\lambda}. \] Using \( h = 6.626\times10^{-34}\ \text{Js},\ c = 3\times10^8\, \text{m/s},\ \lambda = 550\times10^{-9}\ \text{m} \): \[ E = \frac{6.626\times10^{-34}\times3\times10^8}{550\times10^{-9}} = 3.613\times10^{-19}\,\text{J}. \] Converting to electron volts (\(1\,\text{eV}=1.602\times10^{-19}\,\text{J}\)): \[ E = \frac{3.613\times10^{-19}}{1.602\times10^{-19}} = 2.26\,\text{eV}. \]

1. Compare photon energy with work functions:
   • For Cs: \( E = 2.26\,\text{eV} > \phi_{Cs} = 1.9\,\text{eV} \) → Photoelectric effect occurs.
   • For Li: \( E = 2.26\,\text{eV} < \phi_{Li} = 2.5\,\text{eV} \) → No photoelectric emission.
2. Conclusion: Only cesium has a low enough work function for photoemission at this wavelength.

Answer

✅ Option 2: Cs only