**Problem:**

What is the energy change when an electron drops from the n=3 energy state to the 𝑛=1 energy state in a hydrogen atom?

**Solution:**

E = hν = hc/λ

According to the Rydberg formula:

1/λ = R(Z

^{2}/n

^{2}) where

R = 1.097 x 10

^{7}m

^{-1}

Z = Atomic number of the atom (Z=1 for hydrogen)

Combine these formulas:

E = hcR(Z

^{2}/n

^{2})

h = 6.626 x 10

^{-34}J·s

c = 3 x 10

^{8}m/sec

R = 1.097 x 10

^{7}m

^{-1}

hcR = 6.626 x 10

^{-34}J·s x 3 x 10

^{8}m/sec x 1.097 x 10

^{7}m

^{-1}

hcR = 2.18 x 10

^{-18}J

E = 2.18 x 10

^{-18}J(Z

^{2}/n

^{2})

E

_{n=3}

E = 2.18 x 10

^{-18}J(1

^{2}/3

^{2})

E = 2.18 x 10

^{-18}J(1/9)

E = 2.42 x 10

^{-19}J

E

_{n=1}

E = 2.18 x 10

^{-18}J(1

^{2}/1

^{2})

E = 2.18 x 10

^{-18}J

ΔE = E

_{n=3}- E

_{n=1}

ΔE = 2.42 x 10

^{-19}J - 2.18 x 10

^{-18}J

ΔE = -1.938 x 10

^{-18}J

**Answer:**

The energy change when an electron in the n=3 energy state to the n=1 energy state of a hydrogen atom is -1.938 x 10

^{-18}J. The negative value means the atom loses this energy during the transition. This energy is carried away by a photon emitted by the atom.