Official Sample Test question (Physics) (E=hf)

[BWM95]

I understand how to use the E=hf / E=hc/wavelength formula, but this question threw me off because they give you the energy in terms of keV instead of joules and I'm confused as to how they get the correct answer. (I got the question correct, but it was due to an educated guess that ive never seen a x10^- exponent value for frequency.)

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[Asteri]

Hey BMW95,

eV (or keV) is a unit of measurement used in E&M for small quantities of energy (it is much easier to say 1 eV than 1.6x10^(-19) J)
One electron volt is the energy from a elementary charge in a potential field of one volt. Using the equation E = V * q, the energy represented by one electron volt is equal to an elementary charge 1.6x10^(-19) C * 1 V = 1.6x10^(-19) J per electron volt.

Thus, this problem becomes:
140 keV = 140,000 eV * 1.6x10^(-19) J/eV = 2.24x10^(-14) J
Then, using E = hf as you stated,
2.24x10(-14) J / 6.6x10^(-34) = 3.39x10^(19)

Hope this helps!

 

[BWM95]

Hey BMW95,

eV (or keV) is a unit of measurement used in E&M for small quantities of energy (it is much easier to say 1 eV than 1.6x10^(-19) J)
One electron volt is the energy from a elementary charge in a potential field of one volt. Using the equation E = V * q, the energy represented by one electron volt is equal to an elementary charge 1.6x10^(-19) C * 1 V = 1.6x10^(-19) J per electron volt.

Thus, this problem becomes:
140 keV = 140,000 eV * 1.6x10^(-19) J/eV = 2.24x10^(-14) J
Then, using E = hf as you stated,
2.24x10(-14) J / 6.6x10^(-34) = 3.39x10^(19)

Hope this helps!

Yeah that makes sense, I forgot about the E=Vq formula. Thanks alot!