Question

A metal with work function 1.17 eV is illuminated with light of wavelength 437 nm. What stopping potential is required for the photoelectrons

Answer 1

Answer:
`1.67\ V`

Step-by-step explanation:

Given

Work function
`\phi =1.17\ eV`

The wavelength of the light is
`\lambda =437\ nm`

Energy associated with this wavelength is

`E=(hc)/(\lambda)\\\\\Rightarrow E=(1.99* 10^(-25))/(437* 10^(-9))\\\\\Rightarrow E=4.553* 10^(-19)\ J\\\\\Rightarrow E=2.841\ eV`

Stopping potential is

`V=(E-\phi )/(e)\\\\\Rightarrow V=(\left(2.841-1.17\right)e)/(e)\\\\\Rightarrow V=1.67\ V`