Q. Figure represents a graph of kinetic energy of most energetic photoelectrons, K$_{max}$ (in eV), and frequency (v) for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is
A) $1 \times 10^{14}Hz$
B) $1.5 \times 10^{14}Hz$
C) $2.1 \times 10^{14}Hz$
D) $2.7 \times 10^{14}Hz$
Explanation
(d) Given, K$_{max}$ =3 eV
$h = 4.125 \times 10^{- 15}eV$
hv=K$_{max}$ +W
$4.125 \times 10^{- 15} \times 10^{15} = 3 + W$
or $4.125 = \ 3\ + W$
or $W = 1.125$
The threshold frequency
${\ v}_{0} = \frac{W}{h}$
$v_{0} = \frac{1.125}{4.125 \times 10^{- 15}}$
$v_{0} = \frac{1.125 \times 10^{15}}{4.125}$
$v_{0} = 2.72 \times 10^{14}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.