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Light of wavelength lambda(ph)falls on a...

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?

A

For large potential difference `(V gtgt phi//e), lambda_(e)` is approximately halved if V is made four times

B

`lambda_(e)` increases at the same rate as `lambda_(ph)` for `lambda_(ph) lt hc//phi`

C

`lambda_(e)` is approximately halved, if d is doubled

D

`lambda_(e)` decreases with increase in `phi` and `lambda_(ph)`

Text Solution

Verified by Experts

The correct Answer is:
A

`K_("max")= (hc)/lambda_(Ph) - phi`
kinetic energy of `e^(-)` reaching the anode will be
`K=(hc)/lambda_(Ph)- phi+eV`
Now `lambda_(e)=(h)/sqrt(2mK)=(h)/sqrt(2m((hc)/lambda_(Ph)-phi+eV))`
If `eV gt gt phi`
`lambda_(e)=(h)/sqrt(2m((hc)/lambda_(Ph)+eV))`
If `V_(f)=4V_(i)`
`(lambda_(e))_(f)=((lambda_(e))_(i))/(2)`
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