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Step-by-Step Solution
Step 1: Write the general expression for de Broglie wavelength
According to de Broglie’s relation, the wavelength $ \lambda $ of a particle is given by
$$ \lambda = \frac{h}{p} \,, $$
where $ h $ is Planck’s constant and $ p $ is the momentum of the particle. For a particle of mass $ m $, accelerated through a potential difference $ V $, its kinetic energy is $ eV $, and thus
$$ \lambda = \frac{h}{\sqrt{2meV}} \,. $$
Step 2: Use a simplified form for the electron and proton
In many textbooks, the de Broglie wavelength for an electron in Ångström ($ \text{Å} $) units when accelerated through a potential $ V $ (in volts) is often written as:
$$ \lambda_e = \frac{12.27}{\sqrt{V}} \, \text{Å}. $$
For a proton of mass approximately 1836 times that of the electron, the expression can be similarly written (using appropriate constants) as:
$$ \lambda_p = \frac{0.286}{\sqrt{V}} \, \text{Å}. $$
Step 3: Calculate the ratio of their wavelengths
The ratio of the de Broglie wavelength of the electron to that of the proton is:
$$
\frac{\lambda_e}{\lambda_p}
= \frac{\frac{12.27}{\sqrt{V}}}{\frac{0.286}{\sqrt{V}}}
= \frac{12.27}{0.286}.
$$
Step 4: Simplify the ratio
Calculating $ 12.27 / 0.286 $ gives approximately:
$$
\frac{12.27}{0.286} \approx 43.
$$
Step 5: State the final answer
Thus, the ratio of their wavelengths is about
$$ 43 : 1. $$
