Question

The threshold frequency ${\nu }_0$ for a metal is $7.0\times {10}^{14}{s}^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $\nu$ = $1.0\times {10}^{15}{s}^{-1}$

• A. $1.99\times {10}^{-19}J$
• B. $2.99\times {10}^{19}J$
• C. $3.99\times {10}^{-19}J$
• D. $4.99\times {10}^{-19}J$

Answer 1

The correct option is A $1.99\times {10}^{-19}J$

We know that by using Einstein's equation.

Kinetic energy = $\frac{1}{2}{m}_e{V}^2$

= $h(\nu -{\nu }_0)$

=$(6.626\times {10}^{-34}Js)\times (1.0\times {10}^{15}{s}^{-1}-7.0\times {10}^{14}{s}^{-1})$

= $(6.626\times {10}^{-34}Js)\times (10\times {10}^{14}J{s}^{-1}-7.0\times \times {10}^{14}{s}^{-1})$

= $(6.626\times {10}^{-34}Js)\times (3.0\times {10}^{14}{s}^{-1})$

= $1.988\times {10}^{-19}J$