JEE Main Atom PYQ

1.( JEE Main 2022 (Online) 29th June Morning Shift)

$\sqrt{d_{1}}$ and $\sqrt{d_{2}}$ are the impact parameters corresponding to scattering angles 60 $^{\circ}$ and 90 $^{\circ}$ respectively, when an $α$ particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be ____________.

The Crrect Answer is

Impact parameter $\propto$ > $\cot \frac{θ}{2}$

$\Rightarrow \sqrt{\frac{d_{1}}{d_{2}}} = \frac{\sqrt{3}}{1}$

$\Rightarrow d_{1} = 3 d_{2}$

$\Rightarrow x = 3$

2.(JEE Main 2020 (Online) 8th January Morning Slot)

The graph which depicts the results of Rutherford gold foil experiment with $α$ -particales is :
$θ$ : Scattering angle
Y : Number of scattered $α$ -particles detected (Plots are schematic and not to scale)

A. JEE Main 2020 (Online) 8th January Morning Slot Physics - Atoms and Nuclei Question 138 English Option 1

B. JEE Main 2020 (Online) 8th January Morning Slot Physics - Atoms and Nuclei Question 138 English Option 2

C. JEE Main 2020 (Online) 8th January Morning Slot Physics - Atoms and Nuclei Question 138 English Option 3

D. JEE Main 2020 (Online) 8th January Morning Slot Physics - Atoms and Nuclei Question 138 English Option 4

The Correct Answer is Option (C)

From Ernest Rutherford formula. Which indicates that number of particles (Y) that would be deflected by an angle ‘ $θ$ ’ due to scattering is
$Y_{θ} = \frac{K}{\sin^{4} \frac{θ}{2}}$ where K = constant

Option (C) is the correct answer.

3. ( AIEEE 2006)

A. An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge $Z e$ . Then the distance of closest approach for the alpha nucleus will be proportional to

B. $\frac{1}{m}$

C. $\frac{1}{v^{2}}$

D. $\frac{1}{Z e}$

The correct answer is option (C)

Work done to stop the $α$ particle is equal to $K . E .$

$∴$ $q V = \frac{1}{2} m v^{2} \Rightarrow q \times \frac{K (Z e)}{r} = \frac{1}{2} m v^{2}$

$\Rightarrow r = \frac{2 (2 e) K (Z e)}{m V^{2}} = \frac{4 K Z e^{2}}{m v^{2}}$

$\Rightarrow r \propto \frac{1}{v^{2}}$ and $r \propto \frac{1}{m} .$

4. ( AIEEE 2004)

An $α$ -particle of energy $5$ $M e V$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

A. $10^{- 12} c m$

B. $10^{- 10} c m$

C. $1 A$

D. $10^{- 15 a} c m$

The correct answer is option (A)

KEY NOTE :

Distance of closest approach

$r_{0} = \frac{Z e (2 e)}{4 π ε_{0} E}$

Energy, $E = 5 \times 10^{6} \times 1.6 \times 10^{- 19} J$

$∴$ $r_{0} = \frac{9 \times 10^{9} \times (92 \times 1.6 \times 10^{- 19}) (2 \times 1.6 \times 10^{- 19})}{5 \times 10^{6} \times 1.6 \times 10^{- 19}}$

$\Rightarrow r = 5.2 \times 10^{- 14} m = 5.3 \times 10^{- 12} c m$