Question #827e8

1 Answer
Apr 20, 2017

Answer:

#4.09 * 10^(-19)J#

Explanation:

Use the formula:
#E_n = -R_H(1/n^2)#
where :
#R_H = "Rydberg constant" = 2.18 * 10^(-18)J#,
#n = "the energy level"#

For #n = 4#:
#E_4 = -2.18 * 10^(-18)J(1/4^2) = -1.3625 *10^(-19)J#

For #n = 2#
#E_2 = -2.18 * 10^(-18)J(1/2^2) = -5.45 *10^(-19)J#

Then to find the energy of the photon, you just calculate the change in energy.

#DeltaE = E_"initial" - E_"final"#

#DeltaE = (-1.3625 *10^(-19)J) - (-5.45 *10^(-19)J)#

#DeltaE = 4.0875 * 10^(-19)J#

Since you only have 3 significant figures, the answer would be:
#4.09 * 10^(-19)J#