Question 222: Dual Nature of Matter and Radiation

Question

The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2 E$, is $n: 1$. The value of $n$ is $\_\_\_\_$ .
(Assume mass of proton $=$ mass of neutron) $:$

Accepted Answer

The de Broglie wavelength is given by :
$$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $$
Since kinetic energy $\mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$, we can write momentum as $\mathrm{p}=\sqrt{2 \mathrm{mK}}$.
Therefore:
$$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} $$
Let $\mathrm{m}_{\mathrm{p}}$ be the mass of a proton or neutron.
Deuteron (d) consists of 1 proton and 1 neutron. The mass of deuteron is $\mathrm{m}_{\mathrm{d}} \approx 2 \mathrm{~m}_{\mathrm{p}}$
Kinetic Energy $\mathrm{K}_{\mathrm{d}}=\mathrm{E}$
Alpha particle $(\alpha)$ consists of 2 protons and 2 neutrons. The mass of alpha particle is $\mathrm{m}_\alpha \approx 4 \mathrm{~m}_{\mathrm{p}}$
Kinetic Energy $\mathrm{K}_\alpha=2 \mathrm{E}$
The ratio of wavelengths is :
$$ \frac{\lambda_{\mathrm{d}}}{\lambda_\alpha}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{~K}_{\mathrm{d}}}}}{\frac{\mathrm{~h}}{\sqrt{2 \mathrm{~m}_\alpha \mathrm{K}_\alpha}}}=\sqrt{\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{d}}}} 	imes \sqrt{\frac{\mathrm{K}_\alpha}{\mathrm{K}_{\mathrm{d}}}} $$
Substituting the values:
$$ \frac{\lambda_{\mathrm{d}}}{\lambda_\alpha}=\sqrt{\frac{4 \mathrm{~m}_{\mathrm{p}}}{2 \mathrm{~m}_{\mathrm{p}}}} 	imes \sqrt{\frac{2 \mathrm{E}}{\mathrm{E}}}=2 $$
So, the ratio of de-Broglie wavelengths is $2: 1 \Rightarrow \frac{\mathrm{n}}{1}=\frac{2}{1} \Rightarrow \mathrm{n}=2$
Therefore, the value of $n$ is 2 .
Hence, the correct answer is 2.

Dual Nature of Matter and Radiation

The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2 E, is n: 1. The value of n is \_\_\_\_ . (Assume mass of proton = mass of neutron) :

Integer / Numeric Answer Type