For a photoelectron, the frequency is given by the expression :
`v=3.3xx10^(15)((1)/(2^(2))-(1)/(n^(2)))`
If the wavelength of the electron is 6600 Ã…, what will be the value of n ?

Explanation

We know that , `v=c//lambda`
For the electron, `" " lambda =6600 Ã… =6600xx10^(-10)m , c=3xx10^(8)ms^(-1)`
`:' " " v=((3xx10^(8)ms^(-1)))/((6600xx10^(10^(-10)m))=4.54xx10^(14)s`.
According to available information ,
`v=3.3xx10^(15)((1)/(2^(2))-(1)/(n^(2)))or 4.54 xx10^(14)=3.3xx10^(15)((1)/(4)-(1)/(n^(2)))`
`(4.54xx10^(14))/(3.3xx10^(15))=((1)/(4)-(1)/(n^(2))) or 0.137=0.25-(1)/(n^(2))`
`(1)/(n^(2))=0.25-0.137=0.113 or n^(2)=8.85 or n=(8.85)^(1//2)~~3`
Thus, for the photoelectron n=3.