Question

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

• A. $4.5 \times 10^{-7} m$
• B. $4.5 \times 10^{-7} nm$
• C. $4.5 \times 10^{-7} A$
• D. $4.5 \times 10^{-7} cm$

Answer 1

Answer: (A)

$\Delta E =$[Excited state $-$ ground state $]=4.4 \times 10^{-19} J$
According to Plank's quantum theory $\longrightarrow \Delta E =\frac{ nhc }{\lambda}$
$\lambda=\frac{ hc }{\Delta E }$
$=\frac{6.6 \times 10^{-34} J - s \times 3 \times 10^{8} m / s }{4.4 \times 10^{-19} J }$
$=4.5 \times 10^{-7} m$