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Physics · Atomic Physics

JEE Main 2026 — 21 January, Evening Shift — Question 45

A particle having electric charge 3×10−19C3 \times 10^{-19} \mathrm{C} and mass 6×10−27 kg6 \times 10^{-27} \mathrm{~kg} is accelerated by applying an electric potential of 1.21 V . Wavelength of the matter wave associated with the particle is α×10−12 m\alpha \times 10^{-12} \mathrm{~m}. The value of α\alpha is ____\_\_\_\_ . (Take Planck's constant =6.6×10−34 J.s=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s} ) Ans. (10)

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

λ=h2mqV\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}

λ=6.6×10−342×18×10−46×1.21λ=10−11 m=10×10−12 mα=10\begin{aligned} & \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 18 \times 10^{-46} \times 1.21}} & \lambda=10^{-11} \mathrm{~m}=10 \times 10^{-12} \mathrm{~m} & \alpha=10 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Dual Nature of Matter

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