Question

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 $6800A^{\circ }$ is incident on the surface of metal find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not ?
[Velocity of light c = $3\times {10}^{8}m/s$, Planck's constant, $h=6\cdot 63\times {10}^{-34}Js$]

Answer 1

Ginger-Marsden experiment: The setup of the Geiger-Marsden experiment is as shown below.
In this experiment, a narrow beam of $\alpha$-particles from a radioactive source was incident on a gold foil. These 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 n the screen which could be observed through a microscope. This entire setup a 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 $\left(\theta \right)$ of a-particles from its original direction.
They observed that most a-particles passed undeviated and only a few $(\sim 0.14$ scattered by more than$1^o$ Few were deflected slightly and only a few (1 in 8000) deflected by more than ${90}^o$. Some particles even bounced back with 1800.
Mass defect : It is observed that the mass of a nucleus is smaller than the sum of the masses of the constituent nucleons in the free state. The difference between the actual mass of the nucleus and the sum of the masses of constituent nucleons is called mass defect.
The mass defect is
$[Z{m}_p+(A-Z\left)m\right]-M$
where Z is the, atomic number (number of protons), A is the mass number, (A - Z) is the mass of neutrons, $m_p$ is the mass of a proton, m is the mass of a neutron and $M_n$ is the measured mass of a nucleus.
Problems:

Given : ${\varphi }_0=2.3eV=2.3\times 1.6\times {10}^{-19}J=3.68\times {10}^{-19}J$

$\lambda =6800A=6800\times {10}^{-10}m;c=3\times {10}^{8}m/s;$

$h=6.63\times {10}^{-34}Js$

We know that the incident is given as

$v=\frac{c}{\lambda }$

$\therefore v=\frac{3\times {10}^8}{6800\times {10}^{-10}}=4.41\times {10}^{14}Hz$

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

$v_0=\frac{{\varphi }_0}{h}$

$\therefore v_0=\frac{3.68\times {10}^{-19}}{6.63\times {10}^{-34}}=5.55\times {10}^{14}Hz$

Since, $v<v_0$ photoelectron will not be emitted.