Question

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

• A. $1.988\times {10}^{-19}\text{J}$
• B. $4.567\times {10}^{-19}\text{J}$
• C. $1.988\times {10}^{-16}\text{J}$
• D. $4.567\times {10}^{-16}\text{J}$

Answer 1

The correct option is A $1.988\times {10}^{-19}\text{J}$

According to Einstein's equation,
Kinetic energy $=\frac{1}{2}{m}_e{v}^2=h(\nu -{\nu }_0)$
$=(6.626\times {10}^{-34}\text{J s})(1.0\times {10}^{15}{\text{s}}^{-1}-7.0\times {10}^{14}{\text{s}}^{-1})$
$=(6.626\times {10}^{-34}\text{Js})(10.0\times {10}^{14}{\text{s}}^{-1}-7.0\times {10}^{14}{\text{s}}^{-1})$
$=(6.626\times {10}^{-34}\text{Js})(3.0\times {10}^{14}{\text{s}}^{-1})$
$=1.988\times {10}^{-19}\text{J}$