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: Set up the comparison.
An electron (mass $m$) accelerated through $V$ has de-Broglie wavelength $\lambda$. A proton (mass $M$) is accelerated through $9V$. We need its wavelength in terms of $\lambda$.
Step 2: Recall the wavelength of an accelerated charge.
When a charge $q$ is accelerated through potential $V$, its de-Broglie wavelength is $\lambda = \dfrac{h}{\sqrt{2\,m\,q\,V}}$.
Step 3: Note the equal charges.
An electron and a proton carry the same magnitude of charge $e$. With $h$ and $e$ common to both, the wavelength scales as $\lambda \propto \dfrac{1}{\sqrt{mV}}$.
Step 4: Form the proton-to-electron ratio.
\[ \frac{\lambda_p}{\lambda} = \sqrt{\frac{m\cdot V}{M\cdot 9V}} \] The charge $e$ and constant $h$ cancel cleanly.
Step 5: Simplify the radical.
The $V$ cancels, leaving $\dfrac{\lambda_p}{\lambda} = \sqrt{\dfrac{m}{9M}} = \dfrac{1}{3}\sqrt{\dfrac{m}{M}}$, since $\sqrt{9} = 3$.
Step 6: Isolate the proton wavelength.
Multiplying through by $\lambda$: $\lambda_p = \dfrac{\lambda}{3}\sqrt{\dfrac{m}{M}}$.
\[ \boxed{\lambda_p = \dfrac{\lambda}{3}\sqrt{\dfrac{m}{M}}\ \text{(option 1)}} \]