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 surface, then photo-electric effect is possible for the case of: (1) Li only (2) Cs only (3) Neither Cs nor Li (4) Both Cs and Li

Key Concepts

1. Photon Energy Formula
   Every photon has energy $E = \frac{hc}{\lambda}$ where:

   • $h = 6.626\times10^{-34}\,\text{J·s}$ (Planck’s constant)
   • $c = 3.0\times10^{8}\,\text{m/s}$ (speed of light)
   • $\lambda$ is the wavelength in metres

2. Work Function $(\phi)$
   The minimum energy a photon must have to eject an electron from a metal surface.

3. Condition for Photo-Emission
   A photon can eject an electron if $E_{\text{photon}} \geq \phi$

4. Comparing Numbers
   Convert photon energy into the same unit (eV) as the work function, then compare.

Logical Flow to Solve

• Convert the given wavelength (550 nm) to energy using the formula.
• Translate that energy from joules to electron-volts.
• Compare the resulting energy (≈2.25 eV) with each metal’s work function.
• Conclude which metals meet the criterion $E_{\text{photon}} \geq \phi$.

What’s Happening?

Imagine you have two locked doors—one needs a lighter push to open (Cesium) and the other needs a harder push (Lithium). A flashlight beam is like a stream of energy pushes. If the push is strong enough, the door opens and an electron pops out.

Numbers in Simple Words

• Cesium’s lock needs 1.9 energy coins (eV).
• Lithium’s lock needs 2.5 energy coins (eV).
• The yellowish light we shine (550 nm) carries about 2.25 energy coins each.

Who Opens?

• Cesium: needs 1.9 coins, light gives 2.25 → door opens!
• Lithium: needs 2.5 coins, light gives 2.25 → not enough, door stays shut.

So only Cesium lets electrons out.

Step 1: Photon Energy for 550 nm

Convert 550 nm to metres: $\lambda = 550\,\text{nm} = 550\times10^{-9}\,\text{m}$

Calculate energy in joules: $E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{550\times10^{-9}}$ $E \approx 3.61\times10^{-19}\,\text{J}$

Step 2: Convert to Electron-Volts

$1\,\text{eV} = 1.602\times10^{-19}\,\text{J}$ $E = \frac{3.61\times10^{-19}}{1.602\times10^{-19}} \approx 2.25\,\text{eV}$

Step 3: Compare with Work Functions

• Cesium: $\phi_{\text{Cs}} = 1.9\,\text{eV}$
  $2.25\,\text{eV} \ge 1.9\,\text{eV} \;\;\rightarrow\;\; \text{Photo-electron emitted}$
• Lithium: $\phi_{\text{Li}} = 2.5\,\text{eV}$
  $2.25\,\text{eV} < 2.5\,\text{eV} \;\;\rightarrow\;\; \text{No emission}$

Final Answer

Only Cesium exhibits the photo-electric effect.
Option (2) — Cs only