Ncert Exercises & Solutions

A proton and an $α$ -particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths $λ_{p} and λ_{α}$ related to each other?

Hint: The de-Broglie wavelength depends on the momentum given to the particle.

Step 1: Find

λ

in terms of m and q.

\begin{aligned} λ & = \frac{h}{\sqrt{2 mqv}} \\ λ & \propto \frac{1}{\sqrt{mq}} \\ Step 2 : Find the relative wavelengths of proton and α - particle . \\ \frac{λ_{p}}{λ_{α}} & = \frac{{\sqrt{m}}_{α} q_{α}}{\sqrt{m_{p} q_{p}}} = \frac{\sqrt{4 m_{n} \times 2 e}}{\sqrt{m_{ρ} \times e}} = \sqrt{8} \\ λ_{p} & = \sqrt{8} λ_{α} \end{aligned}

i.e wavelength of the proton is

\sqrt{8}

times the wavelength of

α -

particle.

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