Question

Work function \((W_0)\) values of six metals (in eV) are given below:

How many of the above metals do not eject electrons when they are struck with radiation of wavelength 400 nm?

• A. 4
• B. 2
• C. 3
• D. 5

Hint:

For photoelectric effect problems, compare photon energy \(E = \frac{hc}{\lambda}\) with the metal's work function \(W_0\). Electrons are emitted only if \(E \geq W_0\).

Answer 1

The Correct Option is C

Step 1: Calculate the energy of incident photons.
Photon energy is given by \[ E = \frac{hc}{\lambda} \] where \(h = 6.62 \times 10^{-34} \, \text{Js}\), \(c = 3 \times 10^8 \, \text{m/s}\), \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\)
\[ E = \frac{6.62 \times 10^{-34} \cdot 3 \times 10^8}{400 \times 10^{-9}} \approx 4.965 \times 10^{-19} \, \text{J} \] Convert to eV: \[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \implies E \approx 3.1 \, \text{eV} \]

Step 2: Compare photon energy with work function of metals.
Metals eject electrons if \(E \geq W_0\).
Given \(W_0\) values: Li (2.42 eV), Mg (3.7 eV), Cu (4.8 eV), Ag (4.3 eV), K (2.25 eV), Na (2.3 eV)
Photon energy \(E = 3.1 \, \text{eV}\)
- Li: \(2.42 \lt 3.1 \Rightarrow\) electrons emitted - Mg: \(3.7 \gt 3.1 \Rightarrow\) electrons not emitted - Cu: \(4.8 \gt 3.1 \Rightarrow\) electrons not emitted - Ag: \(4.3 \gt 3.1 \Rightarrow\) electrons not emitted - K: \(2.25 \lt 3.1 \Rightarrow\) electrons emitted - Na: \(2.3 \lt 3.1 \Rightarrow\) electrons emitted

Step 3: Count metals that do not emit electrons.
Metals: Mg, Cu, Ag \(\Rightarrow 3\) metals

Step 4: Final conclusion.
Therefore, the number of metals that do not eject electrons is \[ \boxed{3} \]