Question 37: Atoms & Nuclei

Question

The distance of closest approach when an alpha particle of kinetic energy $$6.5 \mathrm{~MeV}$$ strikes a nucleus of atomic number 50 is

Accepted Answer

To find the magnitude of the angular momentum of the electron, we can use the formula for the angular momentum of a particle moving in a circular orbit:
$$L = mvr$$
where:
$$m$$ is the mass of the electron, $$9.1 	imes 10^{-31} \mathrm{~kg}$$
$$v$$ is the speed of the electron, $$2.2 	imes 10^6 \mathrm{~ms}^{-1}$$
$$r$$ is the radius of the orbit, $$0.529 	imes 10^{-10} \mathrm{~m}$$
Now, let's substitute these values into the formula:
$$L = 9.1 	imes 10^{-31} \mathrm{~kg} 	imes 2.2 	imes 10^6 \mathrm{~ms}^{-1} 	imes 0.529 	imes 10^{-10} \mathrm{~m}$$
First, we will calculate the product: $2.2 	imes 10^6 	imes 0.529 	imes 10^{-10}$
$$2.2 	imes 0.529 = 1.1638$$
Combining the exponents of 10:
$$10^6 	imes 10^{-10} = 10^{-4}$$
So the product is:
$$1.1638 	imes 10^{-4}$$
Now calculate:
$$9.1 	imes 10^{-31} \mathrm{~kg} 	imes 1.1638 	imes 10^{-4}$$
Calculating the product of the coefficients: $9.1 	imes 1.1638$
$$9.1 	imes 1.1638 = 10.57758$$
Combining this with the exponent of 10:
$$10.57758 	imes 10^{-31} 	imes 10^{-4} = 10.57758 	imes 10^{-35}$$
We express this final value in scientific notation:
$$1.057758 	imes 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}$$
Rounding this to three significant figures:
$$1.06 	imes 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}$$
Therefore, the correct answer is:
Option A: $$1.06 	imes 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}$$

Answer

0.221 fm

Answer

1.101 fm

Answer

0.0221 fm

Answer

4.42 fm

Atoms & Nuclei

The distance of closest approach when an alpha particle of kinetic energy $6.5 \mathrm{~MeV}$ strikes a nucleus of atomic number 50 is

Choose the correct answer: