Question

If the energy difference between the ground state and excited state of an atom is $4.4x{10}^{-19}J$. The wavelength of photon required to produce this transition is

• A. $3.5x{10}^{-7}m$
• B. $4.5x{10}^{-7}nm$
• C. $3.5x{10}^{-8}m$
• D. $4.5x{10}^{-7}m$

Answer 1

The correct option is D $4.5x{10}^{-7}m$

$\Delta E=4.4\times {10}^{-19}J$
$\Delta E=\frac{hc}{\lambda }$
$\lambda =\frac{6.626\times {10}^{-34}J\times 3\times {10}^{8}m{s}^{-1}}{4.4\times {10}^{-19}J}$
$\lambda =4.5\times {10}^{-7}m$