Question

In a photoelectric effect experiment, the kinetic energy of an emitted electron is $1.986\times 10^{-19}J$, when a radiation of frequency $1.0\times 10^{15}{s}^{-1}$ hits the metal. What is the threshold frequency of the metal (in $s^{-1})$ ?
(Planck's constant $=6.62\times 10^{-34}Js)$

• A. $7.0\times 10^{14}$
• B. $5.8886\times 10^{14}$
• C. $7.0\times 10^{-15}$
• D. $7.0\times 10^{15}$

Answer 1

Answer: (A)

Given, kinetic energy of an emitted electron

$=1.986\times 10^{-19}J$

Frequency of radiation $=1\times 10^{15}{s}^{-1}$

Threshold frequency, $v_0=$ ?

From equation,

$hv-h{v}_0=KE$

$h\left(v-v_0\right)=KE$

$6.62\times 10^{-34}J-s\left(1\times 10^{15}{s}^{-1}-v_0\right)=1.986\times 10^{-19}J$

$1\times 10^{15}{s}^{-1}-v_0=\frac{1.986\times 10^{-19}J}{6.62\times 10^{-34}J-s}$

$v_0=1\times 10^{15}{s}^{-1}-\frac{1.986\times 10^{-19}J}{6.62\times 10^{-34}J-s}$

$v_0=1\times 10^{15}-0.3\times 10^{15}{s}^{-1}$

$v_0=7.0\times 10^{14}{s}^{-1}$