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Mnenie [13.5K]
3 years ago
10

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

Physics
1 answer:
wolverine [178]3 years ago
6 0

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.

Learn more about photons:

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#LearnwithBrainly

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Every object in space exerts a gravitational pull on every other

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Suppose the sphere is electrically neutral. Is it attracted to
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Answer:

No, it is not attracted.

Explanation:

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Answer:

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Explanation:

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Use the formula of time period of simple pendulum

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At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a
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Explanation:

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s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 300 m is the vertical displacement

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Solving for t, we find the it takes for the sack to reach the ground:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

v_x = 60 m/s

We also know the total time of flight,

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Therefore, we can find the horizontal distance travelled by the sack:

d=v_x t = (60)(7.82)=469.2 m

So, the sack will land 469.2 m from the original point below the airplane.

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