Question

what potential difference is needed to give a helium - nucleus (q = 1.2×10^(-19) c) 48 kev of kinetic energy? (list given information, write equations, and answer clearly)

Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that the work done on a charged particle is equal to its change in kinetic energy. For a charged particle moving through a potential difference $V$, the work done $W = qV$, and this is equal to the change in kinetic energy $\Delta K$. Here, the initial kinetic energy $K_i=0$ and the final kinetic energy $K_f = 48\ keV$. So, $W=\Delta K=K_f - K_i=K_f$.

Step2: Identify the charge of a helium nucleus

A helium nucleus has a charge $q = 2e$, where $e=1.6\times10^{-19}\ C$. So $q = 2\times1.6\times10^{-19}\ C=3.2\times10^{-19}\ C$. The kinetic energy $K_f = 48\ keV=48\times10^{3}\ eV$. Since $1\ eV = 1.6\times10^{-19}\ J$, then $K_f=48\times10^{3}\times1.6\times10^{-19}\ J$.

Step3: Solve for the potential difference

From $W = qV$ and $W = K_f$, we can solve for $V$. Rearranging the formula $V=\frac{K_f}{q}$. Substitute $K_f = 48\times10^{3}\ eV$ and $q = 2e$ into the formula. Since $1\ eV$ is the energy gained by a charge of magnitude $e$ when moving through a potential difference of $1\ V$, when $q = 2e$ and $K_f=48\times10^{3}\ eV$, then $V=\frac{48\times10^{3}\ eV}{2e}=24\times10^{3}\ V = 24\ kV$.

Answer:

$24\ kV$