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The de-Broglie wavelength associated with an electron accelerated through a potential difference V is λ. The new de Broglie wavelength of electron when potential difference is increased to 9V will be: |
9λ $\frac{λ}{3}$ 3λ $\frac{λ}{9}$ |
$\frac{λ}{3}$ |
The correct answer is Option (2) → $\frac{λ}{3}$ |
