Question

An $\alpha$ particle has an initial kinetic energy of 25eV and it is accelerated through the potential difference of 150 volt. If a proton has initial kinetic energy of 25eV and it is accelerated through the potential difference of 25 volt then find the approximate ratio of the final wavelengths associated with the proton and the $\alpha$ particle.
a.) 5:1
b.) 13:5
c.) 26:1
d.) 26:5

Answer 1

Hint: To solve this question, we must focus on the wave nature of particles. We should recall the relationship between deBroglie wavelength and kinetic energy of particles.

Complete step by step answer:
We know that Louis deBroglie (Nobel Prize in Physics in 1929) had proposed that a wave function is associated with all particles. Wherever this wave function has non-zero amplitude, we may find the particle in all probability. This means that the intensity of the wave function of a particle at any point is proportional to the probability of finding the particle at that point.
The mathematical relation between kinetic energy of the particle and its deBroglie wavelength can be obtained from the following equation:
$\lambda = \dfrac{h}{p}$
We know, $p = mv$ and $KE = \dfrac{1}{2}m{v^2}$
Momentum is related to kinetic energy as:
$p = \sqrt {2mKE}$
On substituting momentum with kinetic energy, the equation takes the form:
$\lambda = \dfrac{h}{{\sqrt {2m\,KE} }}$
Here, we can see that the wavelength is inversely proportional to mass and kinetic energy of the particle.
The ratio of wavelength therefore:
$\dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{{m_p} \times Energ{y_p}}}{{{m_\alpha } \times Energ{y_\alpha }}}$
Since,
Now we calculate the total kinetic energy.
For alpha particles: Its initial KE was 25eV, its two protons traveled through a potential difference of 150V.
Hence total KE= 25 + 2(150) = 325eV

For proton: Its initial KE was 25eV, it travels through a potential difference of 25eV.
Hence total KE=25 + 1(25) = 50eV
Substituting these values in the formula above we get,
= $\dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \dfrac{{1 \times 50}}{{4 \times 325}} = \dfrac{1}{{26}}$
Hence, the correct answer is Option (C) 26:1

Additional Information:
Although it was proved theoretically before, the first experimental verification of de Broglie's hypothesis came from two physicists working at Bell Laboratories in the USA in 1926. They scattered electrons off Nickel crystals and noticed that the electrons were more likely to appear at certain angles than others. This was the Davisson-Germer experiment.

Note: We must keep in mind the limitation of deBroglie’s principle. DeBroglie’s wavelength is only valid for quantum particles. Although waves are formed for macroscopic particles as well, their wavelength is negligibly small.