Question

The kinetic energy of an electron emitted when a radiation of frequency $1.0\times {10}^{15}{\text{s}}^{-1}$ hits the metal is $2\times {10}^{-19}J$. Calculate the threshold frequency of the metal.

• A. $6.98\times {10}^{14}{s}^{-1}$
• B. $7.52\times {10}^{14}{s}^{-1}$
• C. $3.23\times {10}^{14}{s}^{-1}$
• D. $5.68\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is A $6.98\times {10}^{14}{s}^{-1}$

According to Einstein's equation,
Kinetic energy $=\frac{1}{2}{m}_e{v}^2=h(\nu -{\nu }_0)$
where $m_e$ = mass of electron
$\nu$ = frequency of incident radiation
${\nu }_o$ = Threshold frequency
Given,
Kinetic energy = $2\times {10}^{-19}J$
frequency of the incident radiation $\nu$ = $1.0\times {10}^{15}{\text{s}}^{-1}$

$\therefore 2\times {10}^{-19}=6.62\times {10}^{-34}(1.0\times {10}^{15}-{\nu }_o)$
$\Rightarrow \frac{2\times {10}^{-19}}{6.62\times {10}^{-34}}=1.0\times {10}^{15}-{\nu }_o$
$\Rightarrow {\nu }_o=1.0\times {10}^{15}-0.302\times {10}^{15}$
$\Rightarrow {\nu }_o=0.698\times {10}^{15}{s}^{-1}=6.98\times {10}^{14}{s}^{-1}$