1.

Light of wavelength 0.6mumfrom a sodium lamp fallson a photocell and causes the emission of Photoelectronsfor which the stopping potential is 0.5 V. With light of wavelength 0.4 mum from a sodium lamp, stopping potential is 1.5 V . With this data, the value of h/e is

Answer»

` 4 XX 10^(-19) V s `
`0.25 xx 10^(15) V s `
` 4 xx 10^(-15) V s `
` 4 xx 10^(-8) V s `

Solution :According to Einstein.sphotoelectric equation
` eV= (hc)/(lamda) = phi_(0)`
Where , VStopping POTENTIAL
` lamda ` = Incidentwavelength
` phi_(0)` = Workfunction
` V = ( h/e)c/lamda - (phi_(0))/e`
` V_(1) = ( h/e) c/lamda_(1) = (phi_(0))/e `
` V_(2) = (h/e) c/lamda_(2) - (phi_(0))/e `
Solvingthese two EQUATIONS ,we get
` h/e= (lamda_(1)lamda_(2)) (V_(1)-V_(2))/( c(lamda_(2)-lamda_(1))`
` ((0.6 xx 0.4 xx 10^(-12)) (1.0))/ ( 3xx 10^(8)) (0.2 xx 10^(-6)) = 4 xx 10^(-15) V s `


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