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JEE MainChemistry
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Atomic Structure: JEE Main Chemistry Question with Solution

The wavelength of an electron of kinetic energy 4.50×1029J4.50 \times 10^{-29}\text{J} is...... ×105 m\times 10^{-5}\text{ m}. (Nearest integer) Given : mass of electron is 9×1031 kg9 \times 10^{-31}\text{ kg}, h=6.6×1034 J s\text{h} = 6.6 \times 10^{-34}\text{ J s}
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Question type
Numerical
Exam relevance
JEE Main · Chemistry
Concepts assessed
Chemistry
Academic status
Reviewed by official_key
Source
pyq
Editorial review
9 September 2026

Students also ask

Why is momentum p=2m(KE)p = \sqrt{2m(\text{KE})}?

Since KE=12mv2=p22m\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}, multiplying by 2m2m gives p2=2m(KE)p^2 = 2m(\text{KE}), so p=2m(KE)p = \sqrt{2m(\text{KE})}.