Question 35: Dual Nature of Matter and Radiation

Question

The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_{\mathrm{e}}$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_{\mathrm{p}}$. If their corresponding masses are $m_{\mathrm{e}}$ and $m_{\mathrm{p}}$, respectively, then the ratio of their de Broglie wavelengths $\left(\frac{\lambda_e}{\lambda_p}\right)$ is $\_\_\_\_$ .

Accepted Answer

For a charged particle accelerated through a potential difference $V$, the kinetic energy gained is
$$ K = qV $$
For both electron and proton, the magnitude of charge is the same, that is $e$. So,
$$ K = eV $$
Now, using the non-relativistic expression for kinetic energy,
$$ K = \frac{p^2}{2m} $$
So,
$$ eV = \frac{p^2}{2m} $$
This gives
$$ p = \sqrt{2meV} $$
From de Broglie relation,
$$ \lambda = \frac{h}{p} $$
Substituting $p$,
$$ \lambda = \frac{h}{\sqrt{2meV}} $$
So, for the electron,
$$ \lambda_e = \frac{h}{\sqrt{2m_e eV}} $$
and for the proton,
$$ \lambda_p = \frac{h}{\sqrt{2m_p eV}} $$
Now taking the ratio,
$$ \frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e eV}}{h/\sqrt{2m_p eV}} = \sqrt{\frac{m_p}{m_e}} $$
Hence,
$$ \boxed{\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}} $$
So, the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

$$ 	ext {  } \sqrt{\frac{m_p}{m_e}} $$

Answer

$$ \sqrt{\frac{m_e}{m_p}} $$

Answer

$$ \frac{m_p}{m_e} $$

Answer

$$ \left(\frac{m_p}{m_e}\right)^2 $$

Dual Nature of Matter and Radiation

Choose the correct answer: