The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiation used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength.

Explanation

According to Einstein's rquation, the maximum kinetic energy `E_k` of emitted photoelectrons is given by
`E_(k)=(hc)/(lamda)-W`
where `W=work function.`
let `V_(0)` be the stopping potential. Then, `E_(k)=eV_(0)`
`eV_(0)=(hc)/(lamda)-W` or ` lamda=(hc)/(W+ev_(0))`
or `lamda=((6.62xx10^(-34))(3xx10^(8)))/((1.7+10.4)xx(1.6xx10^(-19)))`
`=1.026xx10^(-7)m=1026A`
This wavelength lies in ultraviolet region. The series lying in ultraviolet region is Lyman series. Hence,
`(1)/(lamda)=R[(1)/((1)^(2))-(1)/(n^(2))]`
`implies(1)/(1.026xx10^(-7))=1.1xx10^(7)[(1)/(1^(2))-(1)/(n^(2))]` or `1-(1)/(n^(2))=(1)/(1.026xx(1.1))`
Solving this equation , we get `n^(2)=9` or `n=3`
Hence, the energy levels involved in hydrogen atom are `n=3` to `n=1`.