Question

Figure represents a graph of kinetic energy of most energetic photoelectrons $K_{\max } \quad$ (in $\left.eV \right)$ 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, $v=10^{15} Hz$,
$K_{\max }=3 eV =3 \times 1.6 \times 10^{-19} \,J$
According to Einstein's photoelectric equation
$K_{\max } =h v-h v_{0} $ or
$ h v_{0}=h v-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}} $
$=(10-7.3) \times 10^{14}=2.7 \times 10^{14} \,Hz$