Light of frequency $8.27 × 10^{14} Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.6 × 10^5 m/s$ are ejected from the surface. The threshold frequency for photoemission of electrons will be (Given: $h= 6.6 × 10^{-34} Js$ and mass of electron = $9 × 10^{-31} kg$)

$5.30 × 10^{14} Hz$

The correct answer is Option (1) → $5.30 × 10^{14} Hz$

Given data:

Frequency of incident light: $f = 8.27 \times 10^{14} \, Hz$

Maximum speed of photoelectrons: $v = 6.6 \times 10^{5} \, m/s$

Planck’s constant: $h = 6.6 \times 10^{-34} \, Js$

Mass of electron: $m = 9 \times 10^{-31} \, kg$

Photoelectric equation:

$hf = hf_0 + K_{max}$

$f_0 = f - \frac{K_{max}}{h}$

Kinetic energy:

$K_{max} = \frac{1}{2}mv^2$

$K_{max} = \frac{1}{2}(9 \times 10^{-31})(6.6 \times 10^{5})^2$

$K_{max} = 0.5 \times 9 \times 10^{-31} \times 4.356 \times 10^{11}$

$K_{max} = 1.96 \times 10^{-19} \, J$

Threshold frequency:

$f_0 = 8.27 \times 10^{14} - \frac{1.96 \times 10^{-19}}{6.6 \times 10^{-34}}$

$f_0 = 8.27 \times 10^{14} - 2.97 \times 10^{14}$

$f_0 = 5.3 \times 10^{14} \, Hz$

Final Answer:

$f_0 = 5.3 \times 10^{14} \, Hz$