gen chem aamc question?

[moldovanits]

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 10–19 J
B 1.4 x 10–25 J
C 4.4 x 10–28 J
D 2.9 x 10–42 J

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[thatscorrect7]

E = hf = hc/(wavelength)

 

[VCU07]

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 10–19 J
B 1.4 x 10–25 J
C 4.4 x 10–28 J
D 2.9 x 10–42 J

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