Question

The distance of closest approach of Alpha particle catering experiment is our if the velocity of Alpha particle is increased by 50% calculate the percentage change

Answer 1

-55.56%

Answer 2

The distance of closest approach ($r_0$) in Rutherford's alpha particle scattering experiment is determined by the conservation of energy. At the distance of closest approach, the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy.

The initial kinetic energy of the alpha particle is $KE = \frac{1}{2}mv^2$. The electrostatic potential energy at the distance of closest approach $r_0$ between an alpha particle (charge $2e$) and a nucleus (charge $Ze$) is $PE = \frac{k(2e)(Ze)}{r_0} = \frac{2kZe^2}{r_0}$, where $k = \frac{1}{4\pi\epsilon_0}$.

By conservation of energy: $\frac{1}{2}mv^2 = \frac{2kZe^2}{r_0}$

From this equation, we can express the distance of closest approach: $r_0 = \frac{4kZe^2}{mv^2}$

This shows that the distance of closest approach is inversely proportional to the square of the velocity of the alpha particle: $r_0 \propto \frac{1}{v^2}$

Let the initial velocity of the alpha particle be $v_1$ and the initial distance of closest approach be $r_1$. $r_1 \propto \frac{1}{v_1^2}$

The velocity of the alpha particle is increased by 50%. So, the new velocity $v_2$ is: $v_2 = v_1 + 0.50v_1 = 1.5v_1$

Let the new distance of closest approach be $r_2$. $r_2 \propto \frac{1}{v_2^2}$

Now, we can find the ratio of the new distance of closest approach to the initial distance of closest approach: $\frac{r_2}{r_1} = \frac{1/v_2^2}{1/v_1^2} = \frac{v_1^2}{v_2^2}$ Substitute $v_2 = 1.5v_1$: $\frac{r_2}{r_1} = \frac{v_1^2}{(1.5v_1)^2} = \frac{v_1^2}{2.25v_1^2} = \frac{1}{2.25}$ To simplify $1/2.25$: $2.25 = \frac{225}{100} = \frac{9 \times 25}{4 \times 25} = \frac{9}{4}$. So, $\frac{r_2}{r_1} = \frac{1}{9/4} = \frac{4}{9}$. This means $r_2 = \frac{4}{9}r_1$.

Now, we calculate the percentage change in the distance of closest approach: Percentage change = $\left( \frac{r_2 - r_1}{r_1} \right) \times 100\%$ Substitute $r_2 = \frac{4}{9}r_1$: Percentage change = $\left( \frac{\frac{4}{9}r_1 - r_1}{r_1} \right) \times 100\%$ Percentage change = $\left( \frac{4}{9} - 1 \right) \times 100\%$ Percentage change = $\left( \frac{4 - 9}{9} \right) \times 100\%$ Percentage change = $\left( -\frac{5}{9} \right) \times 100\%$ Percentage change = $-0.5555... \times 100\%$ Percentage change = $-55.56\%$ (approximately)

The negative sign indicates a decrease in the distance of closest approach.