The proton and α - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and α - particle is (mass of α -particle is 4 times mass of proton,

Question

The proton and α - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and α - particle is (mass of α -particle is 4 times mass of proton, charge of α -particle is 2 times charge of proton)

TestHub · Accepted Answer

The de-Broglie wavelength of a particle accelerated through a potential difference $V$ is given by:

$λ = \frac{h}{2 m q V}$

For a proton:

$λ_{p} = \frac{h}{2 m _{p} q _{p} V}$

For an $α$ -particle:

$λ_{α} = \frac{h}{2 m _{α} q _{α} V}$

The ratio of their de-Broglie wavelengths is:

$\frac{λ _{p}}{λ _{α}} = \frac{m _{α} q _{α}}{m _{p} q _{p}}$

Given $m_{α} = 4 m_{p}$ and $q_{α} = 2 q_{p}$ , substituting these values:

$\frac{λ _{p}}{λ _{α}} = \frac{4 m _{p} \times 2 q _{p}}{m _{p} q _{p}} = 8 = 22$

Answer: $22$

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