CUET · PHYSICS · PYQ PAPER 2023
The de Broglie wavelength of an electron having kinetic energy 56 eV is :
- A 0.022 nm
- B 16.4 \(\text{\AA}\)
- C 0.164 nm
- D 0.0164 nm
Answer & Solution
Correct Answer
(C) 0.164 nm
Step-by-step Solution
Detailed explanation
\(\lambda = \frac{1.226}{\sqrt{E_k}}\text{ nm}\) \(\lambda = \frac{1.226}{\sqrt{56}}\text{ nm}\) \(\lambda \approx 0.164\text{ nm}\)
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