A photelectric meterial having work-function `phi_(0)` is illuminated with light of wavelegth `lamda(lamdalt(hc)/(4_(0))).` The fastes photoelectron has a de Broglie wevelength `lamda_(d).` A change in wavelength of the incident light by `Deltalamda` results in a change `Deltalamda_(d)` in `lamda_(d).` Then the ration `Deltalamda_(d)//Deltalamda` is proportional to

Options

A. `lamda_d^3/lamda^2`
B. `lamda_d^3/lamda`
C. `lamda_d^2/lamda^2`
D. `lamda_d/lamda`

Answer

A

Explanation

From photoelectric equation , where E is the maximum kinetic energy of emitted electron (mass of electron `= m_e`)
`(hc)/lamda=W+E_(max), and p=(hv)/c=h/lamda_d`
`E=p^2/(2m_e)=h^2/(2m_elamda_d^2)`
`(hc)/lamda=phi_0+(h^2)/(2m_elamda_d^2){W=phi_0}`
Differentiate w.r.t `lamda_d`
`(hc)/lamda^2dlamda=h^2/(2m_e)((-2))/lamda_d^3dlamda_d`
`(m_elamda_d^3)/(h^2xxlamda^2).hc=(dlamda_d)/(dlamda)`
`(dlamda_d)/(dlamda) prop (lamda_d^3)/lamda^2`