- #1

ptownbro

- 60

- 0

1. Homework Statement

1. Homework Statement

An electron in the H atom falls from n=(?) to n=3. The wave length of the emitted photon is 1100 nm. From what level did the photon fall?

## Homework Equations

Used the following to get the Energy.

w = c / v

E = h * v

E = h * (c/w)

w = lambda (couldn’t figure out how to use the real lambda symbol)

c = speed of light at 3*10^8 m/s (where “m” is meters and “s” is seconds)

v = frequency in Hz

E = Energy in Joules

h = planks constant at 6.62607*10^-34*J*s (where “J” is joules and “s” is seconds)

Then used the following along with the Energy calculated from above

Change in Energy = -2.18*10^-18 * (1/n2^2 – 1/n1^2)

## The Attempt at a Solution

a) Converted 1100nm to m.

=11*10^-7 mb) Calculated frequency using w = c / v

v = (3*10^8 m/s) / (11*10^-7 m)

v = (3*10^8 s^-1 * 10^7) / 11

v = (3*10^15 s^-1) / 11

v = 2.73*10^14 s^-1c) Calculated Energy using E = h * v (with v calculated in part b)

E = (6.62607*10^-34*J*s) * (2.73*10^14 s^-1)

E = (6.62607*10^-20*J) * (2.73)

E = 18.07*10^-20 J

d) Finally, calculated the level using the Energy

18.07*10^-20 J = (-2.18*10^-18) * (1/3^2 – 1/x^2)

(1/9 – 1/x^2) = (18.07*10^-20 J) / (-2.18*10^-18)

(– 1/x^2) = (18.07*10^-20 J) / (-2.18*10^-18) * 9

x^2 = - (-2.18*10^-18) / (18.07*10^-20 J * 9)

x = [(2.18*10^-18) / (-18.07*10^-20 J * 9)] ^ 1/2

...

Stopped because it didn’t seem right =)