JEE Physics 2026 Question with Solution

Given: electric charge $q = 3 \times 10^{-19} \, \text{C}$, mass $m = 6 \times 10^{-27} \, \text{kg}$, potential difference $V = 1.21 \, \text{V}$, and Planck's constant $h = 6.6 \times 10^{-34} \, \text{J.s}$.

Find: the value of $\alpha$ when the de Broglie wavelength is written as $\alpha \times 10^{-12} \, \text{m}$.

According to the de Broglie hypothesis,

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

For a charged particle accelerated through a potential difference,

$$K = qV$$

and

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

Therefore,

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

Substituting the given values,

$$\sqrt{2mqV} = \sqrt{2 \times (6 \times 10^{-27}) \times (3 \times 10^{-19}) \times 1.21}$$ $$\sqrt{2mqV} = \sqrt{36 \times 1.21 \times 10^{-46}}$$ $$\sqrt{2mqV} = 6 \times 1.1 \times 10^{-23} = 6.6 \times 10^{-23} \, \text{kg}\cdot\text{m/s}$$

Now,

$$\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11} \, \text{m}$$

Write this in the required form:

$$10^{-11} \, \text{m} = 10 \times 10^{-12} \, \text{m}$$

Hence, $\alpha = 10$.

Therefore, the value of $\alpha$ is $10$.

Given: the particle is charged and accelerated through a potential difference.

Find: the associated matter wavelength.

Every moving particle has an associated de Broglie wave, and the wavelength is inversely proportional to momentum. When a charged particle is accelerated through a potential difference, the electrical potential energy gained becomes kinetic energy. Thus the required relation is obtained by combining

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

with

$$qV = \frac{p^2}{2m}$$

This gives

$$p = \sqrt{2mqV}$$

and hence

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

Using the provided values leads to

$$\lambda = 10^{-11} \, \text{m}$$

which is the same as

$$10 \times 10^{-12} \, \text{m}$$

So the required numerical value is $10$.