JEE Physics 2026 Question with Solution

Given: de Broglie wavelengths of a deuteron with kinetic energy $E$ and an alpha particle with kinetic energy $2E$ are to be compared.

Find: the value of $n$ in the ratio $n:1$.

Concept:

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

Thus,

$$\lambda \propto \frac{1}{\sqrt{mK}}$$

Step 1: Determine the masses. A deuteron consists of one proton and one neutron, so

$$m_d = 2m$$

An alpha particle consists of two protons and two neutrons, so

$$m_\alpha = 4m$$

Step 2: Write the wavelength expressions. For the deuteron,

$$\lambda_d \propto \frac{1}{\sqrt{2m \cdot E}}$$

For the alpha particle,

$$\lambda_\alpha \propto \frac{1}{\sqrt{4m \cdot 2E}} = \frac{1}{\sqrt{8mE}}$$

Step 3: Take the ratio.

$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{8mE}{2mE}} = \sqrt{4} = 2$$

Step 4: Express in the required form.

$$n:1 = 2:1$$

So,

$$n = 2$$

Therefore, the value of $n$ is $2$.

Given: $\lambda \propto \frac{1}{\sqrt{mK}}$.

Find: the ratio of wavelengths.

Since constants cancel, compare only $mK$. For the deuteron,

$$mK = 2m \cdot E = 2mE$$

For the alpha particle,

$$mK = 4m \cdot 2E = 8mE$$

Hence,

$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{8mE}{2mE}} = 2$$

Therefore, the required ratio is $2:1$, so the value of $n$ is $2$.