JEE Challenger
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ChemMedAdv 2011 • Paper 1

Calculate Number of Metals Showing Photoelectric Effect at 300 nm

The work function (ϕ)(\phi) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm300\text{ nm} wavelength falls on the metal is

MetalLiNaKMgCuAgFePtWϕ (eV)2.42.32.23.74.84.34.76.34.75\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Metal} & \text{Li} & \text{Na} & \text{K} & \text{Mg} & \text{Cu} & \text{Ag} & \text{Fe} & \text{Pt} & \text{W} \\ \hline \phi\text{ (eV)} & 2.4 & 2.3 & 2.2 & 3.7 & 4.8 & 4.3 & 4.7 & 6.3 & 4.75 \\ \hline \end{array}
Official Numerical Answer4

Topics & Concepts

Structure of AtomBohr Model

Step-by-Step Solution

To determine the number of metals that will show the photoelectric effect, we first calculate the energy (EE) of the incident photons of light with a wavelength of λ=300 nm\lambda = 300\text{ nm}.

The energy of a photon is given by the formula:
E=hcλE = \frac{hc}{\lambda}

Where:

  • h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34}\text{ J}\cdot\text{s} (Planck's constant)
  • c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s} (Speed of light)
  • 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

Substituting the given values:
E=6.626×10−34 J⋅s×3.0×108 m/s300×10−9 mE = \frac{6.626 \times 10^{-34}\text{ J}\cdot\text{s} \times 3.0 \times 10^8\text{ m/s}}{300 \times 10^{-9}\text{ m}}
E=6.626×10−19 JE = 6.626 \times 10^{-19}\text{ J}

Converting this energy into electron-volts (eV\text{eV}):
E=6.626×10−191.6×10−19 eV≈4.14 eVE = \frac{6.626 \times 10^{-19}}{1.6 \times 10^{-19}}\text{ eV} \approx 4.14\text{ eV}

For photoelectric emission to take place, the energy of the incident photon (EE) must be greater than or equal to the work function (ϕ\phi) of the metal:
E≥ϕE \ge \phi

Comparing the photon energy E=4.14 eVE = 4.14\text{ eV} with the given work functions (ϕ\phi):

  1. Li: ϕ=2.4 eV≤4.14 eV\phi = 2.4\text{ eV} \le 4.14\text{ eV} (Photoelectric effect occurs)
  2. Na: ϕ=2.3 eV≤4.14 eV\phi = 2.3\text{ eV} \le 4.14\text{ eV} (Photoelectric effect occurs)
  3. K: ϕ=2.2 eV≤4.14 eV\phi = 2.2\text{ eV} \le 4.14\text{ eV} (Photoelectric effect occurs)
  4. Mg: ϕ=3.7 eV≤4.14 eV\phi = 3.7\text{ eV} \le 4.14\text{ eV} (Photoelectric effect occurs)
  5. Cu: ϕ=4.8 eV>4.14 eV\phi = 4.8\text{ eV} > 4.14\text{ eV} (No photoelectric effect)
  6. Ag: ϕ=4.3 eV>4.14 eV\phi = 4.3\text{ eV} > 4.14\text{ eV} (No photoelectric effect)
  7. Fe: ϕ=4.7 eV>4.14 eV\phi = 4.7\text{ eV} > 4.14\text{ eV} (No photoelectric effect)
  8. Pt: ϕ=6.3 eV>4.14 eV\phi = 6.3\text{ eV} > 4.14\text{ eV} (No photoelectric effect)
  9. W: ϕ=4.75 eV>4.14 eV\phi = 4.75\text{ eV} > 4.14\text{ eV} (No photoelectric effect)

Thus, 44 metals (Li\text{Li}, Na\text{Na}, K\text{K}, and Mg\text{Mg}) will show the photoelectric effect.

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