Light of wavelength `lambda_(ph)`falls on a cathode plate inside a vacuum tube as shown in the figure .The work function of the cathode surface is `phi` and the anode is a wire mesh of conducting material kept at distance d from the cathode. A potential different V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is `lambda_(e) ` which of the following statement (s) is (are) true?

Options

A. `lamda_e` decreases with increases in `phi and lamda_(ph)`
B. `lamda_e` is approximately halved if d is doubled
C. For large potential difference `(V gtgt phi//e),lamda_e` is approximately halved if V is made four times
D. `lamda_e` increases at the same rate as `lamda_(ph) ` for `lamda_(ph) lt hc //phi`

Answer

C

Explanation

Let m be the mass of electron and v be the speed of electron when it reaches anode. The de Broglie wavelength of electron can be written as follows :
`lamda_e=h/(mv)impliesm^2v^2=(h^2)/lamda_e^2implies1/2mv^2=h^2/(2mlamda_e^2)`
`implies` K.E. of electron on reaching anode `=h^2/(2mlamda_e^2)" "....(i)`
Kinetic energy of electron when emitted from cathode
`=(hc)/lamda_(ph)-phi`
K.E. is further increased due to work done by electric field.
`implies` K.E. of electron , on reaching anode
`=(hc)/lamda_(ph)-phi+eV" "...(ii)`
From (i) and (ii), we can write the following :
`h^2/(2mlamda_e^2)=(hc)/lamda_(ph) -phi+eV" "....(iii)`
From (iii) ,we can see that when `phi and lamda_(ph)` are increased then `lamda_e` also increases , hence option (a) wrong .
From (iii) we can see that `lamda_e` is independent of d, hence option (b) is wrong . When V is very large then `h^2/(2mlamda_e^2)~~eV` , hence when V is made 4 times, then `lamda_e` is approximately halved. So option (c) is correct.
From equation (iii) , we can understand that option (d) cannot be correct, hence only option (c) is correct in this equation.