Question

A proton is accelerated through 225 V. Its
de-Broglie wavelength is
(1) 0.1 nm
(2)0.2 nm
(3) 0.3 nm
(4) 0.4 nm

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Answer 1

Answer:

0.2

Explanation:

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Answer 2

Answer:

$\huge{\underline{\overline{\color{darkblue}{\bf{\mid{♡AngelicDoll♡}\mid}}}}}$

➡️ $\bf\green{(2)~0.2~nm}$ ✔️✔️✔️

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$\huge\red{AngelicDoll..!!!}$