The work function of cesium is 2.14 eV. Find (a) the threshold frequency for cesium, and (b) the wavelength of the incident light if the photo current is brought to zero by a stopping potential 0.60 V. Given `h=6.63xx10^(-34)Js`.

Explanation

(a) Given , work function , W = 2.14 eV
Let `v_0` be the threshold frequency . Then
Energy, `hv_0=W`
`implies v_0=W/h`
`=(2.14xx1.6xx10^(-19))/(6.63xx10^(-34))`
`=5.16xx10^(14) Hz`
(b) Given , stopping potential , `V_0=0.60 eV` When photoelectric current becomes zero, then
Maximum kinetic energy of photoelectric
= Potential energy due to stopping potential
`implies K.E._(max)=eV_0`
`implies (hc)/lamda-W=eV_0`
`implies lamda=(hc)/(eV_0+W)`
`=(6.63xx10^(-34)xx3xx10^8)/((0.60+2.14)xx1.6xx10^(-19))`
`=(19.89xx10^(-28))/(2.74xx1.6xx10^(-19))`
`=4.537xx10^(-7)m`
`=4537Å`