When radiation of wavelength 3000 Å is incident on a photosensitive surface, the maximum K.E. of the photoelectrons is 2.5 eV. What is the stopping potential if radiation of wavelength 1500 Å is incident on the same surface ?

Options

A. 4.5 V
B. 5.5 V
C. more than 5.5 V but less than 7V
D. less than 5.5 V but more than 4.5 V

Answer

C

Explanation

`hv=W_(0)+K_(1)`
`:.(hc)/(lambda)=W_(0)+2.5` ...(1)
In the second case, `lambda'=1500 Å=lambda/2`
`:.(hc)/(lambda/2)=W_(0)+K_(2)`
`:.(2hc)/(lambda)=W_(0)+K_(2)` ...(2)
`:.` Subtracting (1) from, (2), we get
`(hc)/(lambda)=K_(2)-2.5`
`:.K_(2)=(hc)/(lambda)+2.5`
`=(6.6xx10^(-34)xx3xx10^(8))/(3000xx10^(-10)xx1.6xx10^(-19))+2.5`
`=4.14eV+2.5 eV=6.64eV`
`:.` The stopping potential `=6.64 V`
`:.` Stopping potential `=6.64 V gt 5 V`