Question

A proton and an alpha particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths (p) is:
$\left( a \right)1$
$\left( b \right)2$
$\left( c \right)\sqrt{8}$
$\left( d \right)\dfrac{1}{\sqrt{8}}$

Answer 1

We have been provided with two particles which are proton and alpha particle. According to De – Broglie, it proposed that the wavelength associated with a particle of mass m moving with speed v is given by $\lambda =\dfrac{h}{p}=\dfrac{h}{mv}.$ So, use this expression. Also, convert the above expression into the other expression which will give you the relation between the wavelength, mass, potential difference and the charge. Calculate the wavelength of the proton and alpha particle separately. Then take the ratio of these two wavelengths and find the value.

Formulas used:
De – Broglie Wavelength is given by
$\lambda =\dfrac{h}{mv}$
where h is the Planck's constant, m is mass and v is velocity.

Complete step by step answer:
We have been given a proton and an alpha particle and both are accelerated through the same potential difference. Now, we need to calculate the ratio of their De – Broglie Wavelength (p).
We know that the expression of De – Broglie Wavelength of a particle of mass m and moving with velocity v is given by,
$\lambda =\dfrac{h}{mv}$
Where, mv = p (momentum)
Consider the proton:
Let ${{m}_{1}}$ be the mass of the proton and ${{k}_{1}}$ be the kinetic energy. Then the De – Broglie Wavelength is given as,
${{\lambda }_{1}}=\dfrac{h}{\sqrt{2{{m}_{1}}{{k}_{1}}}}.....\left( i \right)\left[ \text{Since }p=\sqrt{2{{m}_{1}}{{k}_{1}}} \right]$
where ${{\lambda }_{1}}$ is the De – Broglie wavelength of the proton.
The expression of the kinetic energy k is given as,
${{k}_{1}}=qV.....\left( ii \right)$
where q is the charge of the proton.
Consider the alpha particle:
Let ${{m}_{2}}$ be the mass of the alpha particle and ${{k}_{2}}$ be the kinetic energy, then the De – Broglie Wavelength is given as,
${{\lambda }_{2}}=\dfrac{h}{\sqrt{2{{m}_{2}}{{k}_{2}}}}\left[ \text{Since }p=\sqrt{2{{m}_{2}}{{k}_{2}}} \right]$
where ${{\lambda }_{2}}$ is the De – Broglie wavelength of the alpha particle.
The expression of the kinetic energy of the alpha particle is given as,
${{k}_{2}}={{q}_{0}}V$
where ${{q}_{o}}$ is the charge on the alpha particle.
Also, the alpha particle can be represented as $_{2}^{4}He.$ Then, ${{q}_{o}}=2q$ and ${{m}_{2}}=4{{m}_{1}}.$ Therefore,
${{\lambda }_{2}}=\dfrac{h}{\sqrt{2\times 4{{m}_{1}}\times 2q\times v}}......\left( iii \right)$
And ${{\lambda }_{1}}$ can be written as,
${{\lambda }_{1}}=\dfrac{h}{\sqrt{2{{m}_{1}}qv}}.....\left( iv \right)$
(from (i) and (ii))
Hence, the ratio of their de – Broglie wavelength (p) from equation (iii) and (iv), we get,
$\dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{h}{\sqrt{2{{m}_{1}}qv}}\times \dfrac{\sqrt{2\times {{m}_{1}}\times 4\times 2qv}}{h}$
$\Rightarrow \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{4}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$
$\Rightarrow \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{4}{\sqrt{2}}$
$\Rightarrow \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{2\times 2}{\sqrt{2}}$
$\Rightarrow \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=2\sqrt{2}$
Hence, the ratio of the De – Broglie wavelength of the proton and the de – Broglie wavelength of the alpha particle is $2\sqrt{2}.$
Therefore, option (a) is the right answer.

Note:
The wavelengths of the moving macroscopic objects which are very small like about ${{10}^{34}}m,$ that cannot be measured and we even do not feel their existence. However, the wavelengths of subatomic particles such as the electron are significant and can be measured. The expression, $E=h\nu =\dfrac{hc}{\lambda }$ gives the dual nature of matter or wave.