1.9 × 1015 Hz
1.6×1015 Hz
2 x 1016 Hz
1.2×1015 Hz
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
Match List-I with List-II.
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.
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,