Question

A proton is accelerated through a potential difference of $400V$.To have the same de-Broglie wavelength, what potential difference must be applied across doubly ionised $_8{O^{16}}$atoms.
A.$50volt$
B.$12.5volt$
C.$100volt$
D.None of these

Answer 1

The De Broglie wavelength is a wavelength manifested in all objects in quantum mechanics that determines the probability density of finding the object at a given point in the configuration space, according to wave-particle duality. The momentum of a particle is inversely proportional to its de Broglie wavelength.

Complete answer:
Let us know more about De Broglie wavelength.
In 1924, a French physicist named Louis de Broglie believed that the same relationships apply to particles as they do to photons:
$E = h\nu ,$ $c = \lambda \nu ,$ $E = \dfrac{{hc}}{\lambda } = pc,$
Where $\;E$ and $p$ are the photon's energy and momentum, $\nu$ and $\lambda$ are the photon's frequency and wavelength, $h$ is the Planck constant, and $c$ is the speed of light.
Using the Planck constant and the particle's relativistic momentum, we can define the de Broglie wavelength as follows:
${\lambda _B} = \dfrac{h}{p}$
Unlike photons, which always travel at the same speed, which is equal to the speed of light, the momenta of particles in special relativity are determined by the mass $m$ and the velocity $v$ using the formula:
$p = \dfrac{{mv}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}$
The following is a simplified equation for the de Broglie wavelength:
$\lambda = \dfrac{h}{{mv}}$
Now, let us solve the problem:
As we all know, a particle's de Broglie wavelength is given as
$\lambda = \dfrac{h}{{mv}}$
As a result, for proton, we will have
$KE = qV$
As a result, the photon's momentum is given as
$P = \sqrt {2mK} \\\ P = \sqrt {2mqV} \\\ \\\$
$\therefore \lambda = \dfrac{h}{{\sqrt {2mqV} }}$
Similarly, we will have oxygen now.
Mass of Oxygen$= {m_o} = 16m$
Charge on Oxygen$= Q = 2q$
So we have
$P$ = $\sqrt {{2} \times {16m} \times {2q} \times {V^{'}}}$
Since the wavelength of de Broglie is the same for both, we have
$\dfrac{h}{{\sqrt {64 \times m \times q \times {V^{'}}} }}$ = $\dfrac{h}{{\sqrt {2 \times m \times q \times V} }}$
So we have
$64 \times {V^{'}}$ = $2 \times V$
${V^{'}}$= $\dfrac{V}{{32}}$
${V^{'}}$= $\dfrac{{400}}{{32}}$
${V^{'}}$= $12.5Volts$

Note:
In his thesis, Louis de Broglie proposed that every moving particle, whether microscopic or macroscopic, has a wave character. 'Matter Waves' was the title. He also suggested a relationship between a particle's velocity and momentum and its wavelength if the particle had to behave like a wave.