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NeTakaya
3 years ago
9

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy

of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Physics
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14  + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

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Help me find the natural logarithum of csch^-1 ((-1)/(2 sqrt(30)))
SVEN [57.7K]

The inverse hyperbolic cosecant of anything is a negative number.

In particular, csch⁻¹ [ (-1)/(2√30) ] = csch⁻¹ (≈ -0.0913) = ≈ -3.089

Whatever it is, negative numbers don't have logarithms.

3 0
4 years ago
Which phrases describe all the outer planets motion? Select two options
11111nata11111 [884]

Answer:

slow revolution and  fast rotation

Explanation:

4 0
3 years ago
A bear scratches his claws on a big pine tree. Does the bear do work on the tree? why or why not?
Ganezh [65]

Answer:

See the explanation below.

Explanation:

Work in physics is defined as the product of force by the distance that the body travels in the direction of the force.

It can be represented by means of the following equation.

W=F*d

where:

W = work [J]

F = force [N]

d = distance [m]

In the given example the work is zero since the tree does not move, therefore the bear exerts a force on the tree. But there is no talk of movement of the tree, therefore the work is zero.

8 0
4 years ago
The power of a lens is defined as the reciprocal of its focal length: P = 1/f. (Thus power is measured in inverse meters, called
Len [333]

Answer:

This the power of combination is given by

P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}

Explanation:

Let 'F' be the focal length of the combination of the two lenses and the focal length's of individual lenses be f_{1},f_{2}

We know that focal length is the position of image when object is placed at infinity

Let us place the the object at infinity with respect to the first lens thus the position of image formed by the first lens shall be obtained using lens formula as

\frac{1}{f_{1}}=\frac{1}{u}+\frac{1}{v}

Applying values we get

\frac{1}{f_{1}}=\frac{1}{\infty }+\frac{1}{v}\\\\\Rightarrow \frac{1}{f}=0+\frac{1}{v}\\\\\therefore v=+f_{1}

Now this position of image formed by the first lens act's as object for the second lens, thus we have

\frac{1}{f_{2}}=\frac{1}{-(d-f_{1})}+\frac{1}{v}\\\\\Rightarrow \frac{1}{f_{2}}+\frac{1}{d-f_{1}}=\frac{1}{v}\\\\\therefore \frac{1}{v}=\frac{(d-f_{1})+f_{2}}{f_{2}(d-f_{1})}\\\\=\frac{1}{v}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}\\\\\therefore v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}

Since image of an object placed at infinity will be formed at v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}} thus the focal length of the combination of the 2 thin lenses will be F=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}

This the power of combination is given by

P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}

(For the sake of question we assume lenses to be convex although the same procedure is valid for all other lenses)

5 0
4 years ago
Help!
andrew11 [14]
I think A. Iwbsjdbdjdb
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