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 $\left.=6.62 \times 10^{-34} \,J \,s \right)$

• 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,

$h v-h v_{0}= KE $

$h\left(v-v_{0}\right)= KE$

$6.62 \times 10^{-34}\, J - s \left(1 \times 10^{15} s ^{-1}-v_{0}\right)=1.986 \times 10^{-19}\, J$

$1 \times 10^{15} s ^{-1}-v_{0}=\frac{1.986 \times 10^{-19} J }{6.62 \times 10^{-34} J - s }$

$v_{0}=1 \times 10^{15} s ^{-1}-\frac{1.986 \times 10^{-19} J }{6.62 \times 10^{-34} J - s }$

$v_{0}=1 \times 10^{15}-0.3 \times 10^{15} s ^{-1}$

$v_{0}=7.0 \times 10^{14} \,s ^{-1}$