Question

Question #5cc30

Answer 1

(a)

`sf(KE=5.42xx10^(-19)color(white)(x)J)`

or `sf(3.38color(white)(x)eV)`

(b)

`sf(3.38color(white)(x)V)`

Explanation:

In the photoelectric effect the incident photon strikes an atom and transfers its energy. This energy is used to remove the electron. The excess energy appears as the kinetic energy of the electron.

This can be expressed as :

`sf(E=hf-phi)`

`sf(phi)` is referred to the work function of the metal.

In this case we are given the threshold wavelength `sf(lambda_0)` such that `sf(phi=(hc)/lamda_0)`

`:.` `sf(E=(hc)/lambda-(hc)/lambda_0)`

`sf(E=hc[1/lambda-1/lambda_0])`

`sf(E=(6.63xx10^(-34)xx3.00xx10^(8))/(10^(-9))[1/200-1/440]color(white)(x)J)`

`sf(E=5.42xx10^(-19)color(white)(x)J)`

This is the kinetic energy of the photoelectron.

(b)

In the vaccuum tube these electrons are repelled by the collector plate and the voltage is measured when the current is zero.

If an electron of charge `sf(e)` is moved through a potential difference of `sf(V)` volts then `sf(eV)` Joules of work is done.

`:.` `sf(eV=1/2mv^2=5.42xx10^(-19)color(white)(x)J)`

The electronic charge is `sf(-1.602xx10^(-19)color(white)(x)C)`

`:.` `sf(V=(5.42xxcancel(10^(-19)))/(1.602xxcancel(10^(-19)))=3.38color(white)(x)V)`