Physics Q44 – JEE-MAIN 2026

Home Physics Dual Nature of MatterQuestion #71

The de Broglie wavelength associated with an electron accelerated through a potential difference

V

λ_{e}

, and the de Broglie wavelength associated with a proton accelerated through the same potential difference is

λ_{p}

. If their corresponding masses are

m_{e}

and

m_{p}

, respectively, then the ratio of their de Broglie wavelengths

(\frac{λ_{e}}{λ_{p}})

is _______.

Clue 1 of 3: De Broglie Wavelength

Recall the relationship between de Broglie wavelength, momentum, and kinetic energy for a particle.

🧠 Full Solution Path

Step 1: Relate de Broglie wavelength to kinetic energy✦ Active

The de Broglie wavelength $λ$ is given by $λ = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is momentum. The momentum $p$ can be expressed in terms of kinetic energy $K$ and mass $m$ as $p = \sqrt{2 m K}$ . Combining these, the de Broglie wavelength is $λ = \frac{h}{\sqrt{2 m K}}$ .

Step 2: Express kinetic energy in terms of potential difference○ Expand

When a charged particle with charge $q$ is accelerated through a potential difference $V$ , its kinetic energy gained is $K = q V$ . Substituting this into the de Broglie wavelength equation from Step 1, we get:

λ = \frac{h}{\sqrt{2 m q V}}

Step 3: Calculate the ratio of wavelengths○ Expand

For an electron, the charge is $e$ and mass is $m_{e}$ , so its de Broglie wavelength is $λ_{e} = \frac{h}{\sqrt{2 m_{e} e V}}$ . For a proton, the charge is also $e$ and mass is $m_{p}$ , so its de Broglie wavelength is $λ_{p} = \frac{h}{\sqrt{2 m_{p} e V}}$ . Now, we find the ratio:

\frac{λ_{e}}{λ_{p}} = \frac{\frac{h}{\sqrt{2 m_{e} e V}}}{\frac{h}{\sqrt{2 m_{p} e V}}} = \sqrt{\frac{2 m_{p} e V}{2 m_{e} e V}} = \sqrt{\frac{m_{p}}{m_{e}}}

This matches option 1.

💡 Teacher's Secret Hint

Remember that both electrons and protons have the same magnitude of charge, $e$ .

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: De Broglie Wavelength

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