Question

The threshold frequency $v_0$ for a metal is $7.0\times {10}^{14}se{c}^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency. $v=1.0\times {10}^{15}se{c}^{-1}$.

• A. $4.768\times {10}^{-19}$ Joules
• B. $2.6\times {10}^{-19}$ Joules
• C. $1.98\times {10}^{-47}$ Joules
• D. $1.98\times {10}^{-19}$ Joules

Answer 1

Answer: (D)

$1.98\times {10}^{-19}$ Joules

According to Einstein's photoelectric equation,

Kinetic energy imported to the ejected electron is given by,

$K.E=h\times (\nu -{\nu }_0)$

$\nu =1\times {10}^{15}se{c}^{-1}$ is the frequency of incident light, ${\nu }_0=7\times {10}^{14}se{c}^{-1}$ is the threshold frequency

$K.E$=$6.626\times {10}^{-34}Js$ ($10\times {10}^{14}$ $-$ $7\times {10}^{14}$)$se{c}^{-1}$

$K.E=1.98\times {10}^{-19}J$