Chemistry Q122 – AP-EAMCET 2026

Home Chemistry Dual Nature of MatterQuestion #378

If two particles have same kinetic energy, then the ratio of their de Broglie wavelengths (

λ_{A} : λ_{B}

) is (Given:

m_{A} = 4 m_{B}

)

Clue 1 of 3: De Broglie Wavelength Concept

The de Broglie wavelength describes the wave-like properties of particles and is inversely proportional to the particle's momentum.

🧠 Full Solution Path

Step 1: Recall relevant formulas✦ Active

The de Broglie wavelength ( $λ$ ) of a particle is given by:

λ = \frac{h}{p}

where $h$ is Planck's constant and $p$ is the momentum of the particle. The kinetic energy ( $K E$ ) of a particle is related to its momentum and mass ( $m$ ) by:

K E = \frac{p^{2}}{2 m}

💡 Teacher's Secret Hint

Remember that $p = m v$ , so $K E = \frac{(m v)^{2}}{2 m} = \frac{m^{2} v^{2}}{2 m} = \frac{1}{2} m v^{2}$ . This is a crucial link between momentum and kinetic energy.

Step 2: Express de Broglie wavelength in terms of kinetic energy and mass○ Expand

From the kinetic energy formula, we can express momentum $p$ as:

p = \sqrt{2 m K E}

Substitute this expression for $p$ into the de Broglie wavelength formula:

λ = \frac{h}{\sqrt{2 m K E}}

💡 Teacher's Secret Hint

This derived formula for $λ$ in terms of $m$ and $K E$ is very useful for problems where kinetic energy is constant or given, rather than velocity or momentum directly.

Step 3: Set up the ratio of wavelengths and substitute given conditions○ Expand

We are given that the kinetic energies of the two particles are the same, $K E_{A} = K E_{B} = K E$ . We are also given $m_{A} = 4 m_{B}$ . The de Broglie wavelengths for particles A and B are:

λ_{A} = \frac{h}{\sqrt{2 m_{A} K E}}

λ_{B} = \frac{h}{\sqrt{2 m_{B} K E}}

Now, form the ratio $\frac{λ_{A}}{λ_{B}}$ :

\frac{λ_{A}}{λ_{B}} = \frac{\frac{h}{\sqrt{2 m_{A} K E}}}{\frac{h}{\sqrt{2 m_{B} K E}}} = \sqrt{\frac{2 m_{B} K E}{2 m_{A} K E}} = \sqrt{\frac{m_{B}}{m_{A}}}

Substitute the given mass relationship $m_{A} = 4 m_{B}$ into the ratio:

\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{m_{B}}{4 m_{B}}} = \sqrt{\frac{1}{4}}

💡 Teacher's Secret Hint

Notice how Planck's constant ( $h$ ) and the common kinetic energy ( $K E$ ) cancel out, simplifying the calculation significantly. This is common when dealing with ratios.

Step 4: Calculate the final ratio○ Expand

Calculate the square root to find the ratio:

\frac{λ_{A}}{λ_{B}} = \frac{1}{2}

Thus, the ratio of their de Broglie wavelengths is $1 : 2$ .

💡 Teacher's Secret Hint

Always double-check which ratio is being asked (e.g., A:B vs B:A) to avoid simple inversions. Here it's $λ_{A} : λ_{B}$ .

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Chemistry Question 122 – AP-EAMCET 2026

Clue 1 of 3: De Broglie Wavelength Concept

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