If the work function of a metal is 6.63 eV, then find the threshold frequency for photoelectric effect.
1.9 × 1015 Hz
1.6×1015 Hz
2 x 1016 Hz
1.2×1015 Hz
The Correct Option is B
Solution and Explanation
The threshold frequency \( \nu_0 \) is given by:
\(\phi_0 = h \nu_0\)
**Given:**
- \( \phi_0 = 6.63 \, \text{eV} = 6.63 \times 1.6 \times 10^{-19} \, \text{J} = 1.06 \times 10^{-18} \, \text{J} \),
- Planck’s constant \( h = 6.63 \times 10^{-34} \, \text{J·s} \).
Thus, the threshold frequency is:
\(\nu_0 = \frac{\phi_0}{h} = \frac{1.06 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} \, \text{Hz}.\)
The correct option is (B) : 1.6×1015 Hz
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Concepts Used:
Photoelectric Effect
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Photoelectric Effect Formula:
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,
- h is Planck’s constant.
- ν is the frequency of the incident photon.
- W is a work function.
- E is the maximum kinetic energy of ejected electrons: 1/2 mv².
Laws of Photoelectric Effect:
- The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
- There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
- The maximum kinetic energy of the photoelectrons increases with the increase in the frequency (provided frequency γ > γ Th exceeds the threshold limit) of the incident light. The maximum kinetic energy is free from the intensity of light.
- The process of photo-emission is an instantaneous process.