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 = $9 × 10^{-31} kg$)
Answer & explanation
Correct answer: option 1
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:
$f_0 = 5.3 \times 10^{14} \, Hz$