Answer: [tex]1.67\ V[/tex]
Explanation:
Given
Work function [tex]\phi =1.17\ eV[/tex]
The wavelength of the light is [tex]\lambda =437\ nm[/tex]
Energy associated with this wavelength is
[tex]E=\dfrac{hc}{\lambda}\\\\\Rightarrow E=\dfrac{1.99\times 10^{-25}}{437\times 10^{-9}}\\\\\Rightarrow E=4.553\times 10^{-19}\ J\\\\\Rightarrow E=2.841\ eV[/tex]
Stopping potential is
[tex]V=\dfrac{E-\phi }{e}\\\\\Rightarrow V=\dfrac{\left(2.841-1.17\right)e}{e}\\\\\Rightarrow V=1.67\ V[/tex]
