Question

According to de-Broglie hypothesis if an electron of mass '$m$' is accelerated by potential difference '$V$', the associated wavelength is '$\lambda$'. When a proton of mass '$M$' is accelerated through potential difference $9V$, then the wavelength associated with it is

• A. $\frac{\lambda}{3}\sqrt{\frac{m}{M}}$
• B. $3\lambda\sqrt{\frac{m}{M}}$
• C. $\frac{\lambda}{3}\sqrt{\frac{M}{m}}$
• D. $3\lambda\sqrt{\frac{M}{m}}$

Hint:

To avoid mixing up numerator and denominator terms in a ratio, pull out numeric constant scales first. The proton's voltage is $9\times$ larger, which contributes an immediate $\frac{1}{\sqrt{9}} = \frac{1}{3}$ out front. Since a proton is also much heavier than an electron ($M > m$), its wavelength must be shorter, meaning the smaller mass $m$ stays up top.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to find the de-Broglie wavelength of an accelerated proton relative to the wavelength ($\lambda$) of an electron, given their respective mass values and the distinct acceleration potential differences applied.

Step 2: Key Formula or Approach:
The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:
$$\lambda = \frac{h}{\sqrt{2mqV}}$$ Both an electron and a proton carry the same absolute magnitude of electric charge ($q_e = q_p = e$). Therefore, the wavelength depends inversely on the square root of the product of mass and potential:
$$\lambda \propto \frac{1}{\sqrt{m \cdot V}}$$

Step 3: Detailed Explanation:
Let's write the explicit expressions for both particles:
1. For the electron:
$$\lambda_e = \lambda = \frac{h}{\sqrt{2meV}}$$ 2. For the proton (mass $M$, potential $9V$):
$$\lambda_p = \frac{h}{\sqrt{2M \cdot e \cdot (9V)}} = \frac{h}{\sqrt{18MeV}}$$ Now, let's take the ratio of the proton's wavelength to the electron's wavelength:
$$\frac{\lambda_p}{\lambda} = \frac{\frac{h}{\sqrt{18MeV}}}{\frac{h}{\sqrt{2meV}}} = \sqrt{\frac{2meV}{18MeV}} = \sqrt{\frac{2m}{18M}} = \sqrt{\frac{m}{9M}}$$ Extracting the perfect square number 9 out of the radical sign:
$$\frac{\lambda_p}{\lambda} = \frac{1}{3}\sqrt{\frac{m}{M}}$$ Isolating the final value of $\lambda_p$:
$$\lambda_p = \frac{\lambda}{3}\sqrt{\frac{m}{M}}$$

Step 4: Final Answer:
The de-Broglie wavelength associated with the proton is $\frac{\lambda}{3}\sqrt{\frac{m}{M}}$, which matches option (A).