An α -particle of energy 4 MeV is scattered through 180^o by a fixed uranium nucleus. d

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12.Atoms

medium

An $\alpha$-particle of energy $4\ MeV$ is scattered through $180^o $ by a fixed uranium nucleus. The distance of the closest approach is of the order of

A$1\ \mathring A $

B${10^{ - 10}}\ cm$

C${10^{ - 12}}\ cm$

D${10^{ - 15}}\ cm$

Solution

$r$ is called closet approach
$\Delta \mathrm{K} \cdot \mathrm{E} \cdot=\Delta \mathrm{P} \cdot \mathrm{E}$
$4 \times 10^{6}=\frac{9 \times 10^{9} \times 2 \times 92 \times 1.6 \times 10^{-19}}{r}$
$r=662.4 \times 10^{-16} \mathrm{\,m}$
$=6.62 \times 10^{-14} \mathrm{\,m} \Rightarrow 6.62 \times 10^{-12} \mathrm{\,cm}$

Standard 12

Physics

The following diagram indicates the energy levels of a certain atom when the system moves from $4E$ level to $E$. A photon of wavelength $\lambda _1$ is emitted. The wavelength of photon produced during it's transition from $\frac{7}{3}E$ level to $E$ is $\lambda_2$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be

normal

What was the thickness of the gold foil kept in the Geiger-Marsden scattering experiment?

medium

A beam of fast moving alpha particles were directed towards a thin film of gold. The parts $A',\;B'$ and $C'$of the transmitted and reflected beams corresponding to the incident parts $A, B$ and $C$ of the beam, are shown in the adjoining diagram. The number of alpha particles in

easy

The size of an atom is of the order of

easy

If in Rutherford’s experiment, the number of particles scattered at ${90^o}$ angle are $28$ per min, then number of scattered particles at an angle ${60^o}$ and ${120^o}$ will be

hard