Question

if the energy difference between the ground state to excited state of an atom is 4.4 * 10"-7 J. the wavelenth of proton required to produce this transition is

Answer 1

Answer: $4.5\times 10^{-19}m$

Explanation:

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

$\Delta E=E_2-E_1$ = energy difference

h= planck's constant =$6.6\times 10^{-34}Js$

c= velocity of light =$3.0\times 10^{8}m/s$

Given : $\Delta E= 4.4\times 10^{-7}J$

Thus $4.4\times 10^{-7}=\frac{6.6\times 10^{-34}\times 3.0\times 10^{8}}{\lambda}$

$\lambda=4.5\times 10^{-19}m$