Two positively charged particles $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ have been accelerated across the same potential difference of 200 keV as shown below. ![JEE Main 2026 Chemistry, Structure of Atom — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Structure_of_Atom/6983c1b58bfdbb243563981c/question_1__q_6983c1b58bfdbb243563981c__cdn-question-pool.getmarks.app__28S2_q_74_1__92eaf7b7e1_final_ppt_sync.png) [Given mass of $\mathrm{m}_{1}=1$ amu and $\mathrm{m}_{2}=4$ amu] The deBroglie wavelength of $\mathrm{m}_{1}$ will be $x$ times of $\mathrm{m}_{2}$. The value of $x$ is $\_\_\_\_$ (nearest integer)

JEE Main 2026 — Chemistry Physical Chemistry

2026integereasy

Two positively charged particles $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ have been accelerated across the same potential difference of 200 keV as shown below.

JEE Main 2026 Chemistry, Structure of Atom — question figure

[Given mass of $\mathrm{m}_{1}=1$ amu and $\mathrm{m}_{2}=4$ amu]

The deBroglie wavelength of $\mathrm{m}_{1}$ will be $x$ times of $\mathrm{m}_{2}$ . The value of $x$ is $\_\_\_\_$ (nearest integer)

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For a charged particle accelerated through potential V, the de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mVe}}$ .

The wavelength is inversely proportional to $\sqrt{m}$ .

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

This means $\lambda_1 = 2\lambda_2$ , so x = 2 (nearest integer).

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JEE Main 2026 — Chemistry Physical Chemistry

2026integereasy

Two positively charged particles $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ have been accelerated across the same potential difference of 200 keV as shown below.

JEE Main 2026 Chemistry, Structure of Atom — question figure

[Given mass of $\mathrm{m}_{1}=1$ amu and $\mathrm{m}_{2}=4$ amu]

The deBroglie wavelength of $\mathrm{m}_{1}$ will be $x$ times of $\mathrm{m}_{2}$ . The value of $x$ is $\_\_\_\_$ (nearest integer)

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For a charged particle accelerated through potential V, the de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mVe}}$ .

The wavelength is inversely proportional to $\sqrt{m}$ .

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

This means $\lambda_1 = 2\lambda_2$ , so x = 2 (nearest integer).

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry