Photons of light emitted within the D-lines per second?

[moenste]

Homework Statement

(a) Calculate the energy of one photon of light emitted within the D-lines of a sodium lamp if the wavelength of the D-lines is 589 nm.

(b) In a 200 W sodium street lamp, 30% of input electrical energy is emitted within the D-lines. How many photons of light are emitted within the D-lines per second?

(h = 6.6 * 10^-34 J s, c = 3 * 10⁸ m s^-1)

Answers: (a) 3.4 * 10^-19 J, (b) 1.8 * 10²⁰ s^-1

2. The attempt at a solution
(a) f = c / λ = 3 * 10⁸ / 589 * 10^-9 = 5.09 * 10¹⁴ Hz
E = hf = 6.6 * 10^-34 * 5.09 * 10¹⁴ = 3.4 * 10^-19 J

(b) No idea what to begin with. Any help please?

 

[mfb]

If one apple has a mass of 200 g and I give you 2000 g of apples per second, how many apples per second do you get? How did you calculate that?
What changes if you receive only 30% of the apples I give away?

 

[moenste]

If one apple has a mass of 200 g and I give you 2000 g of apples per second, how many apples per second do you get? How did you calculate that?
What changes if you receive only 30% of the apples I give away?

(i) I get 2,000 / 200 = 10 apples per second.

(ii) It's either 2,000 * .3 = 600 g -> 600 / 200 = 3 apples per second or 10 * .3 = 3 apples per second.

---

200 * 0.3 = 60 W
60 * 3 * 10⁸ = 1.8 * 10¹⁰ s^-1

But the answer is 1.8 * 10²⁰ s^-1.

 

[mfb]

200 * 0.3 = 60 W

Missing units, but the result is right.

60 * 3 * 10⁸ = 1.8 * 10¹⁰ s^-1

The units are wrong, you multiply a power with a speed - the result cannot be a frequency.
Why do you take the speed of light here? It is unrelated to this subquestion.

In general, working with units helps to spot many errors.

 

[moenste]

Missing units, but the result is right.
The units are wrong, you multiply a power with a speed - the result cannot be a frequency.
Why do you take the speed of light here? It is unrelated to this subquestion.

In general, working with units helps to spot many errors.

A lamp supplies 200 W of energy. 30% of this energy is emitted within the D-lines. It is 200 W * 0.3 = 60 W.

Energy of one photon of light emitted within the D-lines = 3.4 * 10^-19 J.

60 W / 3.4 * 10^-19 J = 1.8 * 10²⁰ s^-1.

 

[mfb]

Correct.

Just a notation issue: a/b * c is usually read as (a/b)*c which is not what you meant here.
It is better to put brackets around the denominator: 60 W / (3.4 * 10^-19 J)