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 wavelength `0.4 mum` from a sodium lamp, the stopping potential is 1.5 V. With this data , the value of h/e is

Options

A. `4 xx 10^(-15) V`
B. `3 xx 10^(-15) V`
C. `4 xx 10^(-9) V`
D. `2 xx 10^(-9) V`

Answer

1

Explanation

`(w_(0))/(e)=(hc)/(lambda_(2)e)-V_(01),`
`(w_(0))/(e) = (hc)/(lambda_(2)e)-V_(02)`
`(h)/(c)c[(1)/(lambda_(1))-(1)/(lambda_(2))]=V_(01)-V_(02)`
`rArr(h)/(e)=(V_(01)-V_(02))/(c[(1)/(lambda_(1))-(1)/(lambda_(2))])`