1.

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):}

Answer»

`{:(P,Q,R,S),(3,2,4,1):}`
`{:(P,Q,R,S),(1,3,3,4):}`
`{:(P,Q,R,S),(2,1,3,4):}`
`{:(P,Q,R,S),(4,1,2,3):}`

Solution :`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`


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