# Question 43dbd

Apr 22, 2017

sf(lambda_(n)=(h)/(8sqrt(2meV))

#### Explanation:

I will assume that the charge on the proton is +e and that the alpha particle is 4 times as massive.

The kinetic energy of the proton as it is accelerated through the potential V is given by:

$\textsf{e V = \frac{1}{2} m {v}^{2}}$

$\therefore$$\textsf{{v}^{2} = \frac{2 e V}{m}}$

$\textsf{v = \sqrt{\frac{2 e V}{m}} \text{ } \textcolor{red}{\left(1\right)}}$

The de Broglie wavelength of the proton is given by:

$\textsf{{\lambda}_{p} = \frac{h}{m v} \text{ } \textcolor{red}{\left(2\right)}}$

Substituting the expression for v from $\textsf{\textcolor{red}{\left(1\right)}}$ into $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

sf(lambda_p=h/(msqrt((2eV)/(m)))

The mass of the alpha particle = 4m and the charge = 2e.

$\therefore$sf(lambda_(n)=(h)/(4msqrt((4eV)/(m))

sf(lambda_(n)=h/(4m^(1/2).(4eV)^(1/2))#

$\textsf{{\lambda}_{n} = \frac{h}{4 \times 2 \sqrt{m e V}}}$

$\textsf{{\lambda}_{n} = \frac{h}{8 \sqrt{m e V}}}$