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(B) The number of alpha particles $N(	heta)$ scattered at an angle $	heta$ is given by the relation:$N(	heta) \propto \frac{1}{\sin^4(	heta/2)}$Gi

Vedclass · Accepted Answer

$4 x$

Vedclass · Answer

(B) The number of alpha particles $N(	heta)$ scattered at an angle $	heta$ is given by the relation:$N(	heta) \propto \frac{1}{\sin^4(	heta/2)}$Given:$N(90^{\circ}) = x$We need to find $N(60^{\circ})$:$\frac{N(60^{\circ})}{N(90^{\circ})} = \frac{\sin^4(90^{\circ}/2)}{\sin^4(60^{\circ}/2)}$$\frac{N(60^{\circ})}{x} = \frac{\sin^4(45^{\circ})}{\sin^4(30^{\circ})}$Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$ and $\sin(30^{\circ}) = \frac{1}{2}$:$\frac{N(60^{\circ})}{x} = \frac{(1/\sqrt{2})^4}{(1/2)^4} = \frac{1/4}{1/16} = \frac{16}{4} = 4$Therefore,$N(60^{\circ}) = 4x$.

In Rutherford's experiment,the number of particles scattered at a $90^{\circ}$ angle is $x$ per second. The number of particles scattered per second at an angle of $60^{\circ}$ is:

Similar Questions

How is an atom represented? Why?

$\sqrt{d_{1}}$ and $\sqrt{d_{2}}$ are the impact parameters corresponding to scattering angles $60^{\circ}$ and $90^{\circ}$ respectively,when an $\alpha$-particle is approaching a gold nucleus. For $d_{1} = x d_{2}$,the value of $x$ will be . . . . . .

Choose the correct option from the following options given below.

Which one did Rutherford consider to be supported by the results of experiments in which $\alpha - $ particles were scattered by gold foil?

If an alpha particle with energy $7.7 \text{ MeV}$ is bombarded on a thin gold foil, the closest distance from the nucleus it can reach is . . . . . . $\text{m}$. (Atomic number of gold $= 79$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ in $SI$ units)

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