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The $ \lambda$ of light from the spectral emission line of sodium is $589\: nm$. Find the $KE$ at which an electron would have the same de Broglie wavelength.

$\begin {array} {1 1} (a)\;69 \times 10^{-26} J & \quad (b)\;4.31 \: neV \\ (c)\;3.78 \times 106{-28} J & \quad (d)\;2.36\: \mu\: eV \end {array}$


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Ans : (a)
For the $KE,\: K$ of an electron with velocity v, we have
$K=\large\frac{1}{2}\: m_ev^2$………..(i)
For de Broglie wavelength,
$ \lambda=h/m_ev$
$\Rightarrow \large\frac{v^2 = h^2 }{ \lambda^2m_e^2}$……….(ii)
Substituting eq. (ii) in (i),
$ K = \large\frac{m_eh^2}{2 \lambda^2m_e} \large\frac{h^2}{2\lambda^2m_e}$...........(iii)
$=\large\frac{ (6.6 \times 10^{-34} )^2 }{[2 \times (589 \times 10^{-9} )^2 \times 9.1 \times 10^{-31} ]}$
$≈ 69 \times 10^{-26}\: J$


answered Dec 25, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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