JEE Chemistry 2026 Question with Solution

Given: Both positively charged particles are accelerated through the same potential difference of $200 \, \text{keV}$. Their masses are $m_1 = 1 \, \text{amu}$ and $m_2 = 4 \, \text{amu}$.

Find: The value of $x$ such that the de Broglie wavelength of $m_1$ is $x$ times that of $m_2$.

Concept: De Broglie wavelength is

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$

where the kinetic energy gained is $E = qV$.

Since both particles are accelerated through the same potential difference, they gain the same kinetic energy. Therefore,

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

So,

$$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}}$$

Substituting the given values,

$$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{4}{1}} = 2$$

Hence, $x = 2$.

Therefore, the required value is $2$.

Given: Same accelerating potential for both particles.

Find: The ratio $\lambda_1/\lambda_2$.

For a particle accelerated through a potential difference, the kinetic energy is the same here because the solution states both particles gain the same kinetic energy.

Using

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

with $h$ and $E$ constant for both particles, only mass changes. Thus the wavelength varies inversely as the square root of mass.

So the lighter particle has the larger de Broglie wavelength:

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

and therefore

$$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{4}{1}} = 2$$

Therefore, the de Broglie wavelength of $m_1$ is $2$ times that of $m_2$.