The de Broglie wavelength of an electron of kinetic energy 9 eV is (take h=4×10-15 eV.sec, c=3×1010 cm sec-1 and the mass me of electron as mec2=0.5 MeV)

TS EAMCET · Physics · Dual Nature of Matter

The de Broglie wavelength of an electron of kinetic energy $9 eV$ is (take $h = 4 \times 10^{- 15} eV . \sec$ , $c = 3 \times 10^{10} cm \sec^{- 1}$ and the mass $m_{e}$ of electron as $m_{e} c^{2} = 0.5 MeV$ )

A $4 \times 10^{- 8} cm$
B $3 \times 10^{- 8} cm$
C $4 \times 10^{- 7} cm$
D $3 \times 10^{- 7} cm$

Verified Solution

Answer & Solution

Correct Answer

(A) $4 \times 10^{- 8} cm$

Step-by-step Solution

Detailed explanation

de Broglie wavelength λ=h2mek Given, mec2=0.5Mev me=0.5×1063×10102=5.55×10-16 ⇒ λ=4×10-1525.55×10-169 ∴ λ=4.00×10-8 cm

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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