Question

This is an apparatus for studying the photoelectric effect. The work function of the material beinginvestigated is 3.810–19J. Light of a wavelength 350 nm is incident on the material. What is the lowestvoltage needed between the cathode (C) and the anode (A) to stop any electrons ejected from the cathodefrom reaching the anode?

Answer 1

1.175 V

Step-by-step explanation:

We are given that

Work function of the material=
`w_0=3.8* 10^(-19)` J

Wavelength of light=
`350 nm=350* 10^(-9)`m

`1nm=10^(-9) m`

Energy of photon=Work function+Kinetic energy of electron

`(hc)/(\lambda)=w_0+e\delta V`

Where

`\lambda`=Wavelength of light

`c=3* 10^8 m/s`=Speed of light

`w_0=`Work function

`\delta V`=Small voltage

`h=6.63* 10^(-34) J-s`=Plank's constant

`e=1.6* 10^(-19) C`

Substitute the values then we get

`(6.63* 10^(-34)* 3* 10^8)/(350* 10^(-9))=3.8* 10^(-19)+1.6* 10^(-19)\delta V`

`5.68* 10^(-19)-3.8* 10^(-19)=1.6* 10^(-19)\delta V`

`1.88* 10^(-19)=1.6* 10^(-19)\delta V`

`\delat V=(1.88* 10^(-19))/(1.6* 10^(-19))`

`\delta V=1.175 V`

Hence, the lowest voltage needed between the cathode and the anode to stop any electrons ejected from the cathode from reaching the anode=1.175 V