EBK MODERN PHYSICS
EBK MODERN PHYSICS
4th Edition
ISBN: 9781119495468
Author: Krane
Publisher: VST
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Chapter 3, Problem 1Q
To determine

The energy of photons that would be chosen and the reason.

Expert Solution & Answer
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Answer to Problem 1Q

The energy of the light must be greater than 1.2408×108eV as this energy corresponds to the wavelength of light less than the size of the nucleus.

Explanation of Solution

The diffraction of photons by the nuclei occurs when the wavelength of the light is either equal to or less than the size of the nuclei.

Write the expression for the energy of the photon.

    E>hcλ        (1)

Here, λ is the wavelength of light, h is the plank’s constant and c is the speed of light.

Substitute 4.136×1015eVs for h, 3×108m/s for c, 10×1015m for λ in equation (1).

    E>(4.136×1015eVs)(3×108m/s)(10×1015m)>1.2408×108eV

From above, it is clear that the energy of the photons must be greater than 1.2408×108eV.

Conclusion:

Therefore, the energy of the light must be greater than 1.2408×108eV as this energy corresponds to the wavelength of light less than the size of the nucleus.

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