Question

For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect? (Take Planck’s constant as 6.6 $\times$ 10⁻³⁴ Js)

• A. 50 nm
• B. 100 nm
• C. 150 nm
• D. 200 nm

Hint:

Remember the inverse relationship: Shorter wavelength = Higher energy. If the energy is too low at 200 nm, it will definitely be too low for any wavelength longer than that.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For the photoelectric effect to occur, the energy of the incident photon ($E$) must be greater than or equal to the work function ($\phi_0$) of the metal.

Step 2: Key Formula or Approach:
1. Energy of photon: $E = \frac{hc}{\lambda}$ 2. To simplify, use the conversion: $E (\text{eV}) \approx \frac{1240}{\lambda (\text{nm})}$

Step 3: Detailed Explanation:
Given $\phi_0 = 6.6$ eV. We need to find which wavelength results in an energy $E < 6.6$ eV. 1. For $\lambda = 50$ nm: $E = \frac{1240}{50} = 24.8$ eV (Effect occurs) 2. For $\lambda = 100$ nm: $E = \frac{1240}{100} = 12.4$ eV (Effect occurs) 3. For $\lambda = 150$ nm: $E = \frac{1240}{150} \approx 8.27$ eV (Effect occurs) 4. For $\lambda = 200$ nm: $E = \frac{1240}{200} = 6.2$ eV Since 6.2 eV is less than the work function (6.6 eV), no electrons will be emitted.

Step 4: Final Answer:
200 nm radiation does not give rise to the photoelectric effect.