Find the photon energy in electron volts for electromagnetic waves of the follow

Question

Find the photon energy in electron volts for electromagnetic waves of the following frequencies.

1)(a) 103 MHz in the FM radio band

2)(b) 872 kHz in the AM radio band

2. When a surface is illuminated with electromagnetic radiation of wavelength 531 nm, the maximum kinetic energy of the emitted electrons is 0.54 eV. What is the maximum kinetic energy if the surface is illuminated using radidation of wavelength 390 nm?

3. A neutron in a reactor has kinetic energy of approximately 0.026 eV. Calculate the wavelength of this neutron.
in nm

Explanation / Answer

a) energy is given by

h*frequency

where h=plank's constant

so for 103 MHZ, energy=6.824*10^(-26) J

now 1 eV=1.6*10^(-19) J

hence energy=4.265*10^(-7) eV

b) for 872 kHZ, energy=5.78*10^(-28) J

=3.61*10^(-9) eV

2)energy of wavelength 531 nm=2.339 eV

then work function of the surface=2.339-0.54=1.799 eV

now energy associated with 390 nm=3.1856 eV

so maximum kinetic energy=3.1856-1.799=1.3866 eV

3.as we know,

energy=h*c/wavelength
where c=speed of light

putting the values,
wavleength=4.78*10^(-5) m
=47784.19 nm

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