With the help of a neat labelled diagram, describe the Geiger-Marsden experiment. What is mass defect ?
The photoelectric work function for a metal surface is `2 : 3 eV`. If the light of wavelength `6800 Å` is incident on the surface of metal, find threshold frequency and incident frequency. will there be an emission of photoelectron or not ?
[Velocity of light `c = 3 xx 10^(8) m//s`, Planck's constants, `h = 6.63 xx 10^(-34) Js`]

Explanation

Geiger-Marsden experiment : The setup of the Geiger- Marsden experiment is as shown below .

In this experiment,a narrow beam of `alpha`-particle from a radioactive soure was incident on a gold foil. The scattered `alpha`-particles were detected by the detector fixed on a rotating stand. The detector used had a zinc sulphide screen and a microscope.
The `alpha`-particles produced scintillations on the screen which could be observed through a microscope. This entire setup is enclosed in an evacuated chamber.
They observed the number of `alpha`-particles as a function of scattering angle. Now, the scattering angle is the deviation `(theta)` of `alpha`-particles from its original direction.
They observed that most `alpha`-particles passed undeviated and only a few `(-0.14%)` scattered by more than `1^(@)`. Few were deflected slightly and only a few (1in 8000) deflected by more than `90^(@)`. Some particles even bounced back with `18^(@)`.
Mass defect : It is observed that the mass of a nucleus of smaller than the sum of the masses of the contituent nucleons in the free state. The difference between the actual mass of the nucleus and the sum of the massses of constituent nucleons is called mass defect.
The mass defect is
`Deltam = [Zm_(p) +(A-Z) m_(n)] - M`
where Z is the atomic number (number of protons ), A is the mass number, (A-Z) is the number of neutrons, `m_(p)` is the mass of the a proton, `m_n` is the mass of a neutron and is the measured mass of a nucleus.
Given : `phi_(0) = 2.3 eV = 2.3 xx 1.6 xx 10^(-19) J = 3.68 xx 10^(-19) J`
`lambda = 6800 Å = 6800 xx 10^(-10) m c = 3 xx 10^(8) m//s`.
`h= 6.63 xx 10^(-34) Js`
We know that the incident frequency is given as
`v = c/lambda`
`:. v = (3 xx10^(8))/(6800 xx 10^(-10)) = 4.41 xx 10^(14)`
Now, if the incident frequency is greater than the threshold frequency , then photoelectrons will be emitted from the metal surface. The threshold frequency is given from work function as ltbr `v_(0) = (phi_(0))/(h)`
`:. v_(0) = (3.68 xx 10^(-19))/(6.63 xx 10^(-34)) = 5.55 xx 10^(14) Hz`
Since `v lt v_(0)`, photoelectrons will not be emitted.