M moldovanits Full Member 10+ Year Member 15+ Year Member Jan 27, 2009 #1 Advertisement - Members don't see this ad When the wavelength for maximum light emission of a given scintillator is 450 nm, what is the energy of the light photon? can anyone explain this please? A: 4.4 x 1019 J B 1.4 x 1025 J C 4.4 x 1028 J D 2.9 x 1042 J
Advertisement - Members don't see this ad When the wavelength for maximum light emission of a given scintillator is 450 nm, what is the energy of the light photon? can anyone explain this please? A: 4.4 x 1019 J B 1.4 x 1025 J C 4.4 x 1028 J D 2.9 x 1042 J
T thatscorrect7 Full Member 10+ Year Member 15+ Year Member Jan 27, 2009 #2 E = hf = hc/(wavelength) Upvote 0 Downvote
V VCU07 Member 15+ Year Member Jan 28, 2009 #3 moldovanits said: When the wavelength for maximum light emission of a given scintillator is 450 nm, what is the energy of the light photon? can anyone explain this please? A: 4.4 x 1019 J B 1.4 x 1025 J C 4.4 x 1028 J D 2.9 x 1042 J Click to expand... I got A. E=hf=(6.63x10^-34)(f) c=freq. x wavelength f=c/wavelength =3x10^8/450x10^-9=0.66x10^15 back to origional equation: E=(6.63x10^-34)(0.66x10^15)=4.4x10^-19 Hopefully, that is correct Upvote 0 Downvote
moldovanits said: When the wavelength for maximum light emission of a given scintillator is 450 nm, what is the energy of the light photon? can anyone explain this please? A: 4.4 x 1019 J B 1.4 x 1025 J C 4.4 x 1028 J D 2.9 x 1042 J Click to expand... I got A. E=hf=(6.63x10^-34)(f) c=freq. x wavelength f=c/wavelength =3x10^8/450x10^-9=0.66x10^15 back to origional equation: E=(6.63x10^-34)(0.66x10^15)=4.4x10^-19 Hopefully, that is correct
