Question

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. $2.7\times 10^{14}Hz$

Answer 1

Answer: (D)

From graph, $v=10^{15}Hz$,
$K_{\max }=3eV=3\times 1.6\times 10^{-19}J$
According to Einstein's photoelectric equation
$K_{\max }=hv-h{v}_0$ or
$h{v}_0=hv-K_{\max }$
or $v_0=v-\frac{K_{\max }}{h}$
$=10^{15}-\frac{3\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}$
$=(10-7.3)\times 10^{14}=2.7\times 10^{14}Hz$