MCQ
A proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$
- A$0.1$
- ✓$0.2$
- C$0.3$
- D$0.4$
Energy it gains $=225 \,eV$
$\lambda=\frac{h}{m v}=\frac{\lambda h}{\sqrt{2 m E}}$
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Statement $I$ : Electromagnetic waves are not deflected by electric and magnetic field.
Statement $II$ : The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as $E _0=\sqrt{\frac{\mu_0}{\varepsilon_0}} B_0$
In the light of the above statements, choose the correct answer from the options given below: