De Broglie Wavelength Calculation

The de Broglie wavelength ($\lambda$) is given by the formula:

$ \lambda = \frac{h}{p} $

where $h$ is Planck's constant and $p$ is the momentum.

Momentum ($p$) can be related to kinetic energy ($K$) and mass ($m$) using the formula:

$ K = \frac{p^2}{2m} \implies p = \sqrt{2mK} $

Substituting this into the de Broglie wavelength formula gives:

$ \lambda = \frac{h}{\sqrt{2mK}} $

Particle Properties

We need the properties of the deutron and the alpha particle:

• Deutron: Composed of 1 proton and 1 neutron. Given mass of proton = mass of neutron, let the mass of a proton be $m_p$. Then the mass of the deutron ($m_d$) is approximately $2m_p$. Its kinetic energy ($K_d$) is $E$.
• Alpha Particle: A Helium nucleus ($^4_2He$), composed of 2 protons and 2 neutrons. Its mass ($m_\alpha$) is approximately $4m_p$. Its kinetic energy ($K_\alpha$) is $2E$.

Wavelength Ratio Calculation

Calculate the de Broglie wavelength for the deutron ($\lambda_d$):

$ \lambda_d = \frac{h}{\sqrt{2m_d K_d}} = \frac{h}{\sqrt{2(2m_p)E}} = \frac{h}{\sqrt{4m_p E}} $

Calculate the de Broglie wavelength for the alpha particle ($\lambda_\alpha$):

$ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha K_\alpha}} = \frac{h}{\sqrt{2(4m_p)(2E)}} = \frac{h}{\sqrt{16m_p E}} $

The ratio $\frac{\lambda_d}{\lambda_\alpha}$ is given as $n:1$, which means:

$ \frac{\lambda_d}{\lambda_\alpha} = \frac{n}{1} $

Substitute the expressions for $\lambda_d$ and $\lambda_\alpha$:

$ n = \frac{\frac{h}{\sqrt{4m_p E}}}{\frac{h}{\sqrt{16m_p E}}} = \frac{\sqrt{16m_p E}}{\sqrt{4m_p E}} = \sqrt{\frac{16m_p E}{4m_p E}} = \sqrt{\frac{16}{4}} = \sqrt{4} $

Therefore, $n = 2$. The value $n=2$ lies between 2 and 2.