   Chapter 31, Problem 13PE

Chapter
Section
Textbook Problem

The detail observable using a probe is limited by its wavelength. Calculate the energy of a (−ray photon that has a wavelength of 1 × 10 − 16 m , small enough to detect details about one—tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe.

To determine

The energy of the ? ray photon of wavelength 1×1016m.

Explanation

Given info:

The wavelength of the γ ray photon

λ=1×1016m

Formula used:

The energy of a photon of wavelength λ is given by,

E=hcλ

Here, h is the Planck's constant and c is the speed of light.

Calculation:

Substitute 6.626×1034Js for h, 3.00×108m/s for c and 1×1016m for λ to calculate E.

E=hcλ=6.626× 10 34Js3

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