When light of wavelength 400 nm is incident on the cathode of photocell,the stopping potential recorded is 6 V.If the wavelength of the incident light is increased to 600 nm , calcuulate the new stopping potential.

Explanation

`lamda=400nm = 400xx10^(-9)m`
`lamda = 4xx10^(-7)m`
`V_s=6V`
`lamda=800ņm`
`= 800xx10^(-9)m`
`=8xx10^(-7)m`
Let Stopping Potential corresponding to wavelength of 800 nm is `V._s`
`eV_s=hc(1/lamda-1/lamda_0) .....(1)`
`lamda_0`=Threshold wavelength
`eV_s=hc(1/lamda,-1/lamda_0)` .... (2)
Subtract eqn. (1) from (2)
`eV_s-eV_s=hc[1/lamda,-1/lamda_0-1/lamda+1/lamda_0]`
`e(V_s-V_s)=(hc)/e[1/lamda,-1/lamda]`
`V._s - V_s = (hc)/(e)(1/(lambda.) - 1/(lambda))`
`PV_s-6(6.62xx10^(-34)xx3xx(10)^8)/(1.6xx10^(-9))`
`[1/(8xx10^(-7))-1/(4xx10^(-7))]`
`V_s=6(19.86)/(1.6)xx(10^(-34+8+19))/(10(-7))[1/8-1/4]`
`V_s-6=12.41[(1=2)/8]`
`V._s-6=12.41xx(-1/8)`
`V._s-6=-1.551`
`V._s=-1.1551+6`
`V._s=4.45` volt