A) 
The energy of an x-ray photon used for single dental x-rays is

The energy of a photon is related to its wavelength by the equation

where
is the Planck constant
is the speed of light
is the wavelength
Re-arranging the equation for the wavelength, we find

B) 
The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

Answer:
Transverse wave- Back and forth at right angles to the direction of the wave arrow.
longitudinal wave- bask and forth in the direction of the motion of the motion of the wave.
electromagnetic wave- two alternating waves moving at right angles to each other.
Explanation:
In a longitudinal wave, the particles vibrate at right angles in reference to the wave motion.
In a transverse wave, the particles vibrate parallel to the wave motion
Electromagnetic waves occur as a result of the interaction between two waves and are normally transverse in nature.
Answer:
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