JEE Main · ChemistryHard

$Calculate the ratio of the de Broglie wavelength of an alpha particle λ_{α} to that of a proton λ_{p} if both are accelerated through the same potential difference V .$

A. $$1 : 2$$
B. $1 : 22$
C. $$1 : 4$$
D. $$1 : 1$$

Show correct answer & step-by-step solution

Correct answer: B — $1 : 22$

Solution

$The de Broglie wavelength is given by λ = \frac{h}{2 m q V} .$
$For an alpha particle, m_{α} = 4 m_{p} and q_{α} = 2 e .$
$The ratio is \frac{λ _{α}}{λ _{p}} = \frac{m _{p} q _{p}}{m _{α} q _{α}} = \frac{m _{p} e}{4 m _{p} \cdot 2 e} = \frac{1}{8} = \frac{1}{2 2} .$
$Hence the answer is (B).$

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