Question

An atom absorbs $2eV$ energy and is excited to next energy state. The wavelength of light absorbed will be closest to

• A. ${2000}^{\circ }$
• B. ${4000}^{\circ }$
• C. ${8000}^{\circ }$
• D. ${6000}^{\circ }$

Answer 1

Answer: (D)

${6000}^{\circ }$

Here, an atom absorbs 2eV energy and is excited to next energy state, hence
$\Delta E=2eV=2\times 1.6\times {10}^{-19}J$
Also,
$\Delta E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{\Delta E}$

$\lambda =\frac{(6.6\times {10}^{-34})(3\times {10}^8)}{2\times 1.6\times {10}^{-19}}$

$\lambda =\frac{(19.8\times {10}^{-26})}{3.2\times {10}^{-19}}$

$\lambda =6.1875\times {10}^{-7}$

$\lambda =6187.5A^{\circ }$