Question

A potential difference of 50 volt is applied across an $\alpha$-particle initially at rest

Find (i) K.E acquired by $\alpha$ particle.

(ii) Maximum speed acquired by $\alpha$ particle. $[N_A = 6 \times 10^{23}]$

Answer 1

K.E = $1.6 \times 10^{-17}$ J, v = $6.94 \times 10^4$ m/s

Answer 2

Solution: An $\alpha$-particle is a Helium nucleus, which has a charge of $q = +2e$, where $e = 1.6 \times 10^{-19}$ C is the elementary charge. The mass of an $\alpha$-particle is approximately $m = 4 \text{ amu}$. Using $1 \text{ amu} = 1.66 \times 10^{-27}$ kg, the mass is $m \approx 4 \times 1.66 \times 10^{-27}$ kg $= 6.64 \times 10^{-27}$ kg. A more precise value for the mass of an alpha particle is $6.64465724 \times 10^{-27}$ kg. We will use $m \approx 6.64 \times 10^{-27}$ kg for the calculation.

A potential difference of $V = 50$ volts is applied across the $\alpha$-particle initially at rest. The work done by the electric field on the $\alpha$-particle is $W = qV$. According to the work-energy theorem, the work done is equal to the change in kinetic energy. Since the initial kinetic energy is zero, the kinetic energy acquired is equal to the work done.

(i) K.E acquired by $\alpha$ particle: $K.E. = W = qV$ $K.E. = (2e) \times V$ $K.E. = (2 \times 1.6 \times 10^{-19} \text{ C}) \times (50 \text{ V})$ $K.E. = (3.2 \times 10^{-19} \text{ C}) \times (50 \text{ V})$ $K.E. = 160 \times 10^{-19}$ J $K.E. = 1.6 \times 10^{-17}$ J

The kinetic energy can also be expressed in electron volts (eV): $K.E. = qV = (2e) \times (50 \text{ V}) = 100 \text{ eV}$. In Joules, $100 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 1.6 \times 10^{-17}$ J.

(ii) Maximum speed acquired by $\alpha$ particle: The kinetic energy is related to the speed by the formula $K.E. = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed. $1.6 \times 10^{-17} \text{ J} = \frac{1}{2} \times (6.64 \times 10^{-27} \text{ kg}) \times v^2$ $v^2 = \frac{2 \times 1.6 \times 10^{-17} \text{ J}}{6.64 \times 10^{-27} \text{ kg}}$ $v^2 = \frac{3.2 \times 10^{-17}}{6.64 \times 10^{-27}} \text{ m}^2/\text{s}^2$ $v^2 = \frac{3.2}{6.64} \times 10^{-17 - (-27)} \text{ m}^2/\text{s}^2$ $v^2 = \frac{3.2}{6.64} \times 10^{10} \text{ m}^2/\text{s}^2$ $v^2 \approx 0.4819 \times 10^{10} \text{ m}^2/\text{s}^2$ $v^2 \approx 48.19 \times 10^8 \text{ m}^2/\text{s}^2$ $v = \sqrt{48.19 \times 10^8} \text{ m/s}$ $v = \sqrt{48.19} \times 10^4 \text{ m/s}$ $\sqrt{48.19} \approx 6.942$ $v \approx 6.942 \times 10^4$ m/s

Using a more precise mass $m = 6.64465724 \times 10^{-27}$ kg: $v^2 = \frac{3.2 \times 10^{-17}}{6.64465724 \times 10^{-27}} = \frac{3.2}{6.64465724} \times 10^{10} \approx 0.48161 \times 10^{10} = 48.161 \times 10^8$ $v = \sqrt{48.161} \times 10^4 \approx 6.9405 \times 10^4$ m/s. Rounding to three significant figures, $v \approx 6.94 \times 10^4$ m/s.

The final answer is $\boxed{K.E = 1.6 \times 10^{-17} J, v = 6.94 \times 10^4 m/s}$.

Explanation of the solution: (i) The kinetic energy acquired by the $\alpha$-particle is equal to the work done by the electric field, which is given by $K.E. = qV$. Substitute the charge of the $\alpha$-particle ($q = +2e$) and the potential difference ($V = 50$ V) to calculate the kinetic energy in Joules. (ii) The kinetic energy is also related to the speed by the formula $K.E. = \frac{1}{2}mv^2$. Use the calculated kinetic energy and the mass of the $\alpha$-particle ($m \approx 4$ amu) to solve for the speed $v$.

Answer: (i) K.E acquired by $\alpha$ particle = $1.6 \times 10^{-17}$ J (or 100 eV) (ii) Maximum speed acquired by $\alpha$ particle = $6.94 \times 10^4$ m/s