Determine the Planck's constant h if photoelectrons emitted from a surface of a certain metal by light of frequency `2.2 xx 10^(15)`Hz are fully retarted by a reverse potential of 6.6 V and those ejected by light of frequency `4.6 xx 10^(15)` Hz by a reverse potential of `16.5 eV`.

Explanation

From photo electric effect equation, we have
Here `hv_(1) = phi + eV_(01) ` ….....(1)
and `hv_(2) = phi+ 2eV_(02)` ….........(2)
Subtracting equation ( 1) from equation ( 2) ,we get
`h ( v_(2) - v_(1)) = e ( v_(02) - v_(01))`
or
`h = ((v_(02)-v_(01))( 1.6 xx 10^(-19)))/( ( v_(2) - v_(1)))`
or
`h- ((16.5 - 6.6) ( 1.6 xx 10^(-19)))/( ( 4.6 -2.2 ) xx 10^(15))`
or `= 6.6 xx 10^(-34) J-s`