Question

In a photoelectric effect experiment, irradiation of a metal with light of frequency $5.2\times {10}^{14}{sec}^{-1}$ yields electrons with maximum kinetic energy $1.3\times {10}^{-19}J$. Calculate the $v_0$ of the metal.

• A. $3.2\times {10}^{14}{sec}^{-1}$.
• B. $1.6\times {10}^{14}{sec}^{-1}$.
• C. $6.4\times {10}^{14}{sec}^{-1}$.
• D. $3.2\times {10}^{15}{sec}^{-1}$.

Answer 1

The correct option is A $3.2\times {10}^{14}{sec}^{-1}$.

$KE=h(v-v_0)$

Where $v=$ incident frequency,

$v_0=$ threshold frequency.

$\therefore v_0=v-\frac{KE}{h}=5.2\times {10}^{14}-\frac{1.3\times {10}^{-19}}{6.626\times {10}^{-34}}=3.24\times {10}^{14}se{c}^{-1}$