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 x 10^-19 J
• B. 2.99 x 10¹⁹ J
• C. 3.99 x 10^-19 J
• D. 4.99 x 10^-19 J

Answer 1

The correct option is A

1.99 x 10^-19 J

We know that by using Einstein's equation.

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

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

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

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

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

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