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A deuteron moving with one tenth of velocity of light has a certain de Broglie wavelength of $λ .$ An alpha particle having certain kinetic energy has the same de-Broglie wavelength $λ .$ The ratio of kinetic energy of deuteron and that of alpha particles is:

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4 : 1

1 : 4

1 : 2

2: 1

answer is D.

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Detailed Solution

$K = \frac{D^{2}}{2 m}, D = \frac{h}{λ}$

$\begin{array}{l} ∴ K = \frac{h^{2}}{2 m λ^{2}} \\ K α \frac{1}{m} \\ \frac{K_{D}}{K_{α}} = \frac{m_{λ}}{m_{D}} = \frac{4 m}{2 m} \\ = \frac{2}{1} \\ ∴ K_{D} : K_{α} = 2 : 1 \end{array}$

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