A photon of frequency n causes photoelectric emmission from a surface with threshold frequency `n_0` .The de Broglie wavelength `lambda ` of the photoelectron emitted is given as

Options

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

Explanation

According to the photoelectric equation
`E_(1) = 1E +KE`
or `E_(1)` = Theshould or work function `+KE`
`hv= hv_(0) + (1)/(2) mv^(2)`
`rArr (1)/(2)mv^2) = h(v - v_(0)) = Deltav `…..(i)
We know that
`lambda = (h)/(p) = (h)/(mv)`
`:. v= (h)/(m lambda)`
Substinuting this value in equation (i) , we get
`(1)/(2)m(h^(2))/(m^(2)lambda^(2)) = h Delta v `
`(h)/(2 m lambda^(2)) = Delta v `
`:. lambda = sqrt((h)/(2m DElta v`
Hence the answer is d