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

Answer

a

Explanation

If `(KE)_(i)` is the KE of electron just after ejection from cathode, then
`(KE)_(i)=(hc)/(lambda_(ph))-phi`
Final kinetic energy of electron while reaching anode is
`(KE)_(f)=(KE)_(i)+eV=((hc)/(lambda_(ph))-phi)+eV`
If `V gt gtphi//e` or `eVgt gtphi`, then `(KE)_(i)~~eV`
or `1/2mv_(max)^(2)=eV` or `mv_(max)=sqrt(2meV)`
De-Broglie wavelength,
`lambda_(e)=h/(mv_(max))=h/(sqrt(2meV))` or `lambda_(e) prop1/(sqrt(V))`
If `V` is made `4V`, then
`(lambda'_(e))/(lambda_(e))=sqrt(V/(4V))=1/2` or `lambda'_(e)=(lambda_(e))/2`
Thus option (a) is true.
If `phi` and `lambda_(ph)` increases, then `(KE)_(f)` decreases.
But `(KE)_(f)=eV`, so `V` decreases, hence `lambda_(e)` increases.
Thus option (b) is correct.
Also `lambda_(e)=h/(sqrt(2m(KE)_(f)))` or `(dlambda_(e))/(dt)ne(dlambda_(ph))/(dt)`
Thus, option (c) is incorrect.
`lambda_(e)` is independent of `s`, hence option (d) is incorrect.