Photons of energy 5 eV and incident on the surface of metal `M_1` and electrons with kinetic energy `E_(M_1)` and of energy 5.75 eV are incident on surface of metal `M_2` , the electrons with kinetic energy `E_(M_2)=(E_(M_1)+2eV)` and de Broglie wavelength `lamda_(M_2)=(lamdaM_(1))/3` are ejected. If `phi_(M_1) and phi_(M_2)` are work function of metal surface , `M_1 and M_2` respectively, then choose the correct option /s.
`{:(,"List -I",,"List - II"),(P,E_(M_1),1,2.25eV),(Q,E_(M_2),2,0.25eV),(R,phi_(M_1),3,4.75eV),(S,phi_(M_2),4,3.50eV):}`

Options

A. `{:(P,Q,R,S),(3,2,4,1):}`
B. `{:(P,Q,R,S),(1,3,3,4):}`
C. `{:(P,Q,R,S),(2,1,3,4):}`
D. `{:(P,Q,R,S),(4,1,2,3):}`

Answer

C

Explanation

`E_(M_1)=5 eV -phi_(M_1)" "...(i)`
`E_(M_2)=5.75eV-phi_(M_2)" "... (ii)`
Given, `E_(M_2)=E_(M_1)-2eV " "...(iii)`
Also given , `lamda_(M_1)=3lamda_(M_2)" ":.lamda=K/sqrtE`
`E=K/lamda^2`
`K/sqrtE_(M_1)=3K/sqrt(E_(M_2))impliesE_(M_2)=9E_(M_1)" "...(iv)`
From equations (iv) and (iii)
`8E_(M_(1))=2eV`
`E_(M_(1))=0.25eV`
`E_(M_(2))=2.25eV`
`phi_(M_(1))=4.75eV`
`phi_(M_2)=3.50 eV`