Question

The following 3 items relate to a 50,000-watt radio station that transmits 750 kHz(WSB).

1. Determine the number of joules transmitted per second.
A. 750
B. 50 000
C. 750, 000
D. 1.6x10^19
E. 8x10^23

3. calculate the energy of a single photon at the transmitted frequency.
A. 5.0x10^-31 j
B. 1.4x10^-29 j
C. 5.0x10^9-28 j
D. 1.4x10^-19 j
E. 7.5x10^-5 j

2. what is the wavelength of the radio waves emitted from this radio station in meters per cycle.( hint : use the wave equation)
A. 1.4x10^-29
B. 5
C. 10.5
D. 75
E. 400

Answer 1

1 watt power=1 joule work per second

Answer 2

the energy of a photon E=hf where h=plank's constant and f=frequency

Answer 3

I don't get it

Answer 4

first one?

Answer 5

yes

Answer 6

if i do 1 joule work in one second...then you can say my power is 1 Watt...

Answer 7

similarly the power of radio station is 50,000 Watt that means it transmitted 50,000 joules per second

Answer 8

okay thank you , number 2 please ?

Answer 9

E=hf

Answer 10

the energy of a photon E=hf where h=plank's constant and f=frequency

Answer 11

@smartG Are you on the second one - number 3?

Answer 12

"calculate the energy of a single photon at the transmitted frequency."
I'll quote souvik, as his last comments answer that. These comments:
"\(E=hf\)"
and
"the energy of a photon \(E=hf\) where \(h\)=plank's constant and \(f\)=frequency"
Planck's constant is \(6.62606957 × 10^{-34}\ m^2\ kg / s\).

Answer 13

so you just plug it in to he formula ?

Answer 14

yeah

Answer 15

how do I know what to plug in for what ?

Answer 16

As Eric wrote h=6.6*10^-34 SI unit
f=750,000 Hz

Answer 17

i didn't get one of the answer choices

Answer 18

it is 4.9*10^-28
~5*10^-28

Answer 19

so # 2 is c?

Answer 20

I got \[E=hf\]\(6.636\times10^{-31}\ \ \times\ \ 750000\ \ =\ \ 4.9695 \times 10^{-28}\)
http://www.wolframalpha.com/input/?i=6.626e-34+*+750e3
If you round this to two significant figures, you find your answer.

Answer 21

Might have been a math error.

Answer 22

As for the next problem...
The wave equation it speaks of is \(v=f\lambda\), where
\(v\) is the velocity at which the wave propagates,
\(f\) is the frequency, and
\(\lambda\) is the wavelength you seek.
So, you can rearrange that, solving for \(\lambda\).

Do you know the velocity of a electromagnetic wave?
(Same for all electromagnetic waves, like radio waves, light, ultraviolet radiation, and other names for other frequencies).
You'll get it! :)