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# An electron is 2000 times lighter than a proton. Both are moving such that their matter waves have a length of $1\: \dot {A}$. The ratio of their kinetic energy in approximation is

$\begin {array} {1 1} (a)\;1:1 & \quad (b)\;1: 2000 \\ (c)\;2000:1 & \quad (d)\;1:200 \end {array}$

Can you answer this question?

Ans : (c)
Since both have same de Broglie wavelength,
hence both must have equal value of momentum.
Since $E =\large\frac{ p^2}{ pm}$
$\Rightarrow \large\frac{E_e}{ E_p} = \large\frac{m_p}{ m_e} = 1840 = 2000$
{nearest possible approximation to answer}

answered Dec 27, 2013
edited Mar 13, 2014