A photon of frequency n causes photoelectric emmission from a surfece with thereshold .The de Broglie wavelength `lambda ` of the photoelectrn emitted is given as

Options

A. `Delat n = (h)/(2m lambda)`
B. `Delat n = (h)/(lambda)`
C. `[(1)/(v_(0)) - (1)/(v)] = (mc^(2))/(h)`
D. `lambda = sqrt((h)/(2m Delta n))`

Answer

D

Explanation

`E_(1)= IE + KE`
or
`E_(1) = `Threshold E (or) work function `+ KE`
`(hv = hv_(0) + (1)/(2) m u^(2))`
or
`hn = hn_(0) + (1)/(2) m u^(2)`
`(1)/(2) m u^(2) = h(n - n_(0)) = h Delta n`…….(i)
`(lambda = (h)/(mu) ,:. U = (h)/(m lambda))`
Subsitute the value n in equation (i)
`(1)/(2) m(h^(2))/(m^(2)lambda^(2)) = h DRelta n`
`(h)/(2lambda^(2)m) = Delta n `
`:. lambda = sqrt((h)/(2m Delta n))`