NEETPhysicsDual Nature of Radiation and MatterMCQ+4 / −1
The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:
- A13.6 nm
- B136 nm
- C1.36 nm
- D0.136 nm
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Correct answer: D
$${\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o = {{12.27} \over {\sqrt {81} }}\mathop A\limits^o = {{12.27} \over 9}\mathop A\limits^o $$
$$ = 1.36\mathop A\limits^o $$ ($\because$ $$1\mathop A\limits^o = {1 \over {10}}$$ nm)
$ = 0.136$ nm
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