A metal has work function $ 3{\text{ ev}} $ which was exposed to radiation associated with $ 300{\text{ }}{{\text{A}}^ \circ } $ wavelength .Find out
 $ \left( i \right){\text{ K}}{\text{.}}{{\text{E}}_{\max }} $ $ \left( {ii} \right){\text{ }} $ Stopping potential

Hint :Use the concept of photoelectric effect, where the total energy of an incident photon is always equal to the sum of work function and maximum kinetic energy of the photon. The maximum kinetic energy of an emitted electron is directly proportional to the frequency of the incident light.
Energy $ \left( E \right) $ $ = {\text{ }}\dfrac{{hc}}{\lambda } $

Complete Step By Step Answer:
Given parameters:
 Wavelength $ \left( \lambda \right) $ = $ 300{\text{ }}{{\text{A}}^ \circ } $
Work function = $ 3{\text{ ev}} $
From the concept of photo emission we came to know that energy of incident photo electrons is used to eject electrons from metal only when it is greater than the work function of metal and this extra energy is kinetic energy which helps in ejection of electrons from metal. Due to which an electron ejects out the moving electron thus possesses kinetic energy.
 $ \left( i \right) $ Maximum energy $ = {\text{ }}\dfrac{{hc}}{\lambda }{\text{ - work function}} $ $ \_\_\_\left( 1 \right) $
Here, $ h{\text{ = }} $ Planck constant $ = {\text{ 6}}{\text{.626 }} \times {\text{ 1}}{{\text{0}}^{^{ - 34}}}{\text{ J - s}} $
 $ {\text{c = }} $ speed of light $ = {\text{ 3}} \times {\text{ 1}}{{\text{0}}^8}{\text{ m/s}} $
Using value of hc in electron volt :
 $ hc{\text{ = 12400 ev}} $ $ \left( {{A^ \circ }} \right) $
On putting the value in $ \left( 1 \right) $ we get:
 $ K.{E_{_{\max }}} $ $ = {\text{ }}\dfrac{{12400}}{{300}}{\text{ - 3 ev}} $
 $ K.{E_{\max }}{\text{ = 38}}{\text{.3 ev}} $
 $ K.{E_{\max }}{\text{ = 38}}{\text{.3 }} \times {\text{ 1}}{\text{.6 }} \times {\text{ 1}}{{\text{0}}^{ - 19}}{\text{ J}} $
 $ K.{E_{\max }}{\text{ = 6}}{\text{.12 }} \times {\text{ 1}}{{\text{0}}^{ - 18}}{\text{ J}} $
Thus we can calculate kinetic energy of electrons with the use of work function and energy of incident radiation or photons. Also we can find the threshold frequency .
 $ \left( {ii} \right) $ Stopping potential is the maximum energy per electron . Stopping potential is also known as cut-off potential. It is that potential which makes the value of kinetic energy equal to zero.
Therefore,
Stopping potential $ = {\text{ }}\dfrac{{maximum{\text{ energy}}}}{{electron}} $
Stopping potential $ = {\text{ 38}}{\text{.3 }}\dfrac{{ev}}{e} $
Stopping potential $ = {\text{ 38}}{\text{.3 v}} $ .

Note :
The value of $ hc{\text{ = 1240 ev }} $ only when wavelength is in nanometer , otherwise for Armstrong it is $ hc{\text{ = 12400 ev }} $ . Make sure use of units while solving. Kinetic energy can be calculated both in electron volts and joules. Stopping potential increases as energy of incident photons increases.