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Light of wavelength 0.6 mum from a so...

Light of wavelength 0.6 ` mum ` from a sodium lamp falls on a photocell and causes the emission of Photoelectrons for 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

A

` 4 xx 10^(-19) V s `

B

`0.25 xx 10^(15) V s `

C

` 4 xx 10^(-15) V s `

D

` 4 xx 10^(-8) V s `

Text Solution

Verified by Experts

The correct Answer is:
C
According to Einstein.s photoelectric equation
` eV= (hc)/(lamda) = phi_(0)`
Where , V Stopping Potential
` lamda ` = Incident wavelength
` phi_(0)` = Work function
` 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 `
Solving these 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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