The wavelength of light from the spectral emission line of sodium is 662 nm. The kinetic energy at which an electron would have same de-Broglie wavelength would be: $(h=6.62× 10^{-34} J-s, m_e=9×10^{-31} kg)$

$5.5 × 10^{-25} J$

The correct answer is Option (1) → $5.5 × 10^{-25} J$

The De-Broglie wavelength of a light (λ) and its kinetic energy is related by -

$λ=\frac{h}{\sqrt{m_eK.E.}}$

where,

$λ=662nm=662×10^{-9}m$

$m_e$ = mass of electron = $9×10^{-31}kg$

$⇒K.E.=\frac{h^2}{2m_eλ^2}$

$=\frac{(6.62×10^{-34})^2}{2×9×10^{-31}×(662×10^{-9})^2}$

$=5.5 × 10^{-25} J$