An excited atom emits a photon of red light with a wavelength of 732. nm, what is the energy of the photon?

Answer 1

You would use the equations

E(Energy)= h(Planks constant)*Frequency(v)

frequency(v) = c(speed of light)/wavelength(lambda)

Now if you substitute

E=h*(c/lambda)
E*lambda=hc
E=(hc)/lambda

Energy=[(6.626x10^-34 J*sec)*(3.00x10^8m/s)]
........................7.32 x 10^-7 m

Energy=2.72x10^-19Joules

Answer 2

E = hv

c = L x v

so substitute

E = h(c/L)

E = energy = unknown
h = Planck's constant = 6.63 x 10^-34 J*s
c = speed of light = 3.00 x 10^8 m/s
L = wavelength = 732 nm = 732 x 10^-9 m

All units will cancel out except J

**Please Note**
I am a teacher and I like to lead you to the answer - not work out the problem for you and give you the answer. I feel people learn better by an explanation than they do from being given the answer.

Answer 3

E = hv

c = L x v

so substitute

E = h(c/L)

E = energy = unknown
h = Planck's constant = 6.63 x 10^-34 J*s
c = speed of light = 3.00 x 10^8 m/s
L = wavelength = 732 nm = 732 x 10^-9 m

All units will cancel out except J

Answer 4

capability of a photon = (plank's consistent)(frequency) considering we've the wavelength.....we are able to get the frequency with the aid of dividing the value of sunshine with the aid of the photon's wavelength. Wavelength = 717nm x (10^-9 m/ 1nm) = 7.17 x 10^-7m Freq = (3.00 x 10^8) / (7.17 x 10^-7) = 4.18 x 10^14 now that we've the frequency of the photon, you merely could multiply it with the aid of plank's consistent... that's (6.626 x 10^-34 J-s) x (4.18 x 10^14) = capability of photon = 2.seventy seven x 10^-19 Joules

Answer 5

e = hv
v= c/wavelength = 3x10^8/ 732x10^-9 = 4.1x10^14 Hz
e = 4.1x10^14(h)
I forgot what h was (6.636 x 10^-31 i think)