Question

A nucleus in a transition from an excited state emits a gamma-ray photon with an energy of 2.5 MeV. (a)

Answer 1

a) The frequency of the photon is $7.16\cdot 10^{20}Hz$

b) The wavelength of the photon is $4.19\cdot 10^{-13} m$

c) The wavelength of the photon is about 100 times larger than the nuclear radius

Explanation:

a)

The energy of a photon is given by

$E=hf$ (1)

where:

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

f is the frequency of the photon

The photon in this problem has an energy of

$E=2.5 MeV = 2.5\cdot 10^6 eV$

And keeping in mind that

$1eV = 1.6\cdot 10^{-19} J$

we can convert to Joules:

$E=(2.5\cdot 10^6)(1.9\cdot 10^{-19})=4.75\cdot 10^{-13} J$

And now we can use eq.(1) to find the frequency of the photon:

$f=\frac{E}{h}=\frac{4.75\cdot 10^{-13}}{6.63\cdot 10^{-34}}=7.16\cdot 10^{20}Hz$

b)

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

$c=f\lambda$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

f is the frequency

$\lambda$ is the wavelength

For the photon in this problem,

$f=7.16\cdot 10^{20}Hz$

Re-arranging the equation, we find its wavelength:

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8}{7.16\cdot 10^{20}}=4.19\cdot 10^{-13} m$

c)

The size of the nuclear radius is approximately

$d \sim 10^{-15} m$

While we see that the wavelength of this photon is

$\lambda=4.19\cdot 10^{-13} m$

Therefore, the ratio between the wavelength of the photon and the nuclear radius is

$\frac{\lambda}{d}=\frac{\sim 10^{-13}}{\sim \cdot 10^{-15}}=100$

So, the wavelength of the photon is approximately a factor 100 times larger than the nuclear radius.

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